tag:blogger.com,1999:blog-4107303980700785197.comments2013-09-08T16:13:57.398-04:00Mr. Andrade's Physics ClassesDavid Andradenoreply@blogger.comBlogger26125tag:blogger.com,1999:blog-4107303980700785197.post-19514073093919849112010-05-18T09:29:27.846-04:002010-05-18T09:29:27.846-04:00Boronny Touch, Elizabeth Trejo, Roneisha Handy
...Boronny Touch, Elizabeth Trejo, Roneisha Handy <br /><br /><br />The four forces involved in rocket flight are weight, thrust, and the aerodynamic forces, lift and drag. The magnitude of the weight depends on the mass of all of the parts of the rocket. There are many different types of rockets. They are compressed air rockets, water rockets, model rockets, and full scale rockets. During the entire flight of a compressed air rocket only the aerodynamic forces and weight act on the rocket. Unlike compressed air rockets, the mass of the bottle rocket varies during flight. There are four systems in a full scale rocket, the structural system, payload system, guidance system, propulsion system.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-71348995857581045582010-03-22T09:42:48.760-04:002010-03-22T09:42:48.760-04:00Eunice Polanco, Cynthia Garcia, Courtney Manns, Do...Eunice Polanco, Cynthia Garcia, Courtney Manns, Dondre Edwards, Angelo Faiella<br /><br />The area around an object where another mass will experience gravitational force is known as a gravitational field. <br /><br />When field lines are close together, the force is greater on the mass.<br /><br />G is the gravitational constant. Its value is 6.67 x 10 ^ -11 Nm squared kg to the negative 2.<br /><br />Magnitude and direction of the field strength, along with the acceleration due to gravity are always equal, or the same.<br /><br />At any point on the surface, the direction of the field is always vertically down.<br /><br />The gravitational field strength is nearly constant in magnitude when closer to the surface of the Earth which is g=9.81 N kg^-1.<br /><br />If an object of mass m2 is traveling with speed v in a circular orbit of radius r around a planet of mass m1 then the centripetal force F=m2v2 /r is equal to the gravitational force on the object F=Gm1m2/r2<br /><br />For all circular orbits round a particular planet of mass m1, there is only one orbit with any specific valve of speed, radius or period.<br /><br />A geostationary satellite remains above a fixed point on the Earth’s equator with an orbital period of 24hrs.<br /><br />The gravitational field is the area round an object where another mass will experience a gravitational force. <br /><br />The field is represented by field lines which show the direction of the force on a mass in the field. <br /><br />The closer together is the field lines, the stronger the force on the mass. <br /><br />Gravitational field strength at any point is proportional to the mass (m1) of the object causing the field, and inversely proportional to the square of the distance r to the point from the centre of the object. This can be written as g = Gm1/r2 <br /><br />The force on the mass (m2) in the field is given by F = gm2, which gives F = Gm1m2/r2<br /><br />The gravitational potential at any point in the field is given by V = -Gm1/r<br /><br />G is the gravitational constant and its value is 6.67 x 10-11 N m2 kg-2<br />Close to the surface of the Earth the gravitational field strength is nearly constant in magnitude, g = 9.81 N kg-1<br /><br />The magnitude and direction of the field strength, and the acceleration due to gravity is always the same. <br /><br />The direction of the field is vertically downwards at any point on the surface. <br /><br />Because the variation in the gravitational field strength is negligible, the potential energy gained by an object moving in the gravitational field can be taken to be directly proportional to the increase in height. <br /><br />ΔEp = mgΔh<br /><br />If an object of mass m2 is traveling with speed v in a circular orbit of radius r around a planet of mass m1, then the centripetal force F = m2v2/r is equal to the gravitational force on the object F = Gm1m2/r2<br /><br />For all circular orbits round a particular plant of mass m1, there is only one orbit with any specific value of speed, radius or period<br /><br />A geostationary satellite remains above a fixed pint on the Earth’s equator with an orbital period of 24 hours.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-80002312994519695122010-03-22T09:41:40.789-04:002010-03-22T09:41:40.789-04:00Margarita Rodriguez
Victoria Taiwo
Kevin D.
Aldrin...Margarita Rodriguez<br />Victoria Taiwo<br />Kevin D.<br />Aldrin M.<br /><br /> The gravitational field is the round area of an object where another mass will experience a gravitational force. Field lines show the direction of the forceon a mass on the field represents the field. Then when the force on the mass becomes stronger and this is because the field lines are closer together. Gravitational force can be found g=F/m. Where F is force and m is mass. The gravitational constant can be written as G and the constant is 6.67*10^-11 N m ^2 kg^-2. Acceleration due to gravity is always the same to magnitude and direction. The direction of the field is vertically downwards at any point on the surface. The centripetal force is can be written in the as F=m2v^2/r. Where F is force, m is mass,v is the velocity, and r is the radius. The grvitational force can be written as F=Gm1m2/r^2. Where G is the gravitational constant, m1 is the first mass, m2 is the second mass, r is the radius. <br /><br />~It is surpring that the centripetal force is equal to the gravitational force.-Margarita<br /><br />~I found it surprising that for all circular orbits round a particular planet mass m1, there is only one orbit with any specific value of speed, radius or period.-Kevin<br /><br />~I think it is interesting that the feild is represtented by feild lines which shows the direction of the force on a mass in the feild!-Victoria<br /><br />~I found it surprising how a geostationary satellite remains above a fixed point on the Earth's equator with an orbital period of 24 hours- AldrinAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-72617089721029290432010-03-22T09:41:40.790-04:002010-03-22T09:41:40.790-04:00Boronny Touch
Sergio Munoz
Allen Ngu
Adrian Simps...Boronny Touch <br />Sergio Munoz<br />Allen Ngu<br />Adrian Simpson<br /><br /><br />In this lesson, we learned that gravitational force is directed towards the center of the planet. The closer the field lines are to each other, the stronger the gravitational force. The field is the same at the same distance from the center of the planet whether the body is extended or compressed. We have potential energy because we do work against gravity to get to a certain point. An object field strength always decreases as the height increase, since the field lines of the Earth’s surface it always directed downwards and it never varies with the height. There is a very small percentage of the field strength in addition we also learned that the field strength is measure in force per unit mass. The gravitational field strength and the acceleration of gravity is the same all the time. The gravitational potential energy gains energy as the height of the field lines are increasing. An object that travels in circular orbit has a centripetal force (F= m2v2) that is equal to the gravitational force (F=Gm1m2/r2) on the objects. All objects in circular orbit around a certain planet of mass, only has any specific value of speed, radius or period. The geostationary satellite remains a certain point above the earth’s equator with a 24 hour period of orbit. In conclusion we learned a series of formulas. For example, delta Ep=mg delta h, which shows the correlation between the potential energy, mass, acceleration due to gravity, and the change in height. We also learned that G is the gravitational constant is equal to 6.67 × 10–11 N m2 kg–2.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-43038642998199714102010-03-22T09:24:34.231-04:002010-03-22T09:24:34.231-04:00Elizabeth Trejo
Phyllisse Lewis
Roneisha Handy
Pre...Elizabeth Trejo<br />Phyllisse Lewis<br />Roneisha Handy<br />Precious Humphrey<br /><br />We learned that the gravitational field is the area around an object where another mass would experience a gravitational force. The closer field lines are to each other, the stronger the force on the mass. The formula for gravitational field strength is g= F/m. F stands for the gravitational force and m stands for the mass. The gravitational field strength at any point on the field is equal to the mass of the object causing the field, which can be written as g= Gm/r2. The universal constant of gravitation, G, is 6.67 x 10-11 N m2 kg-2 . Close to the earth’s surface, the gravitational field strength is approximately equal to 9.8 N kg-1. When an object travels in a circular orbit around a planet of mass, the gravitational force on the object can be figured out by using the equation: F= Gm1m2/r2.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-64205277913126122842010-03-22T08:45:23.973-04:002010-03-22T08:45:23.973-04:00Sherwin Yu, Emmanuel Torres
We found out that the ...Sherwin Yu, Emmanuel Torres<br />We found out that the gravitational field is the area between a object and another mass that expirences a gravitational force.Gravitational field strength at any point is proportional to the mass m1of the object causing the field. If a rocket goes too fast it will fly out of earth's orbit and if it's too slow it will go crashing back down to Earth.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-39096769294189622222010-03-22T08:45:23.974-04:002010-03-22T08:45:23.974-04:00Alyssa Zargos, Kevin Andrade, Mohammed Hamid
One...Alyssa Zargos, Kevin Andrade, Mohammed Hamid<br /><br /><br />One of Isaac Newton's great achievements was his formula for the universal law of gravitation. Newton's universal law of gravitation can be explained with the equation F = Gm1m2/r2, where F is the force between the two objects, m1 and m2 are their masses and r is the distance between them. The gravitational field is the area around an object where another mass will experience a gravitational force. There is a force exerting on a mass in the field which is represented by field lines. The closer together the field lines, the stronger the force on a mass. The gravitational force per unit mass is the gravitational field strength. If you get really far away from the Earth the force of gravity on you gets to be nothing and you would no longer be pulled towards the Earth, so you no longer have energy that can be changed back to another form. Gravitational potential V is calculated from the equation V = –Gm1/r, where G is the gravitational constant, m1 is the mass of the planet and r is the distance from the centre of the planet. The potential energy of a mass m2 at any point is given by Ep = m2V.<br /><br />The field close to the surface of the Earth, is the gravitational field strength that is almost constant in magnitude, g = 9.81 Nkg-1. The gravitational field strength and the acceleration due to gravity always have the same value. The field strength varies, and as it decreases as the height increases. Potential energy gained depend on the mass of the object is proportional to the mass. The increase in gravitational potential energy is directly proportional to the increase in height and mass. This can be written ΔEp = kmΔh, where k is a constant of proportionality, m is the mass and Δh is the increase in height.<br /><br /> Centripetal force is F = m2v2/r and the force of gravity F = Gm1m2/r2. Those two equations are equal to each other if an object of mass m2 is travelling with speed v in a circular orbit of radius r around a planet of mass m1. In the multimedia picture, the spacecraft orbited around the earth. As the speed increases, the rocket goes into an elliptical orbit which takes it further from the Earth. The speed needed to keep the spacecraft orbiting around earth is 3073 m s–1. For all circular orbits round a particular planet of mass m1, there is only one orbit with any specific value of speed, radius or period. Law of gravitation, gravitational field strength, gravitational potential energy, and centripetal force is very important when dealing with the force of objects and more specifically objects in space.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-6270792192466036832010-03-22T08:40:40.847-04:002010-03-22T08:40:40.847-04:00Marissa Martinez
Naelia Barragan
Nefertiti Ferguso...Marissa Martinez<br />Naelia Barragan<br />Nefertiti Ferguson-Goforth<br />Lizzette Martinez<br /><br />Newton's universal law of gravitation can be summarized with the equation F = Gm1m2/r2, where F is the force between the two objects, m1 and m2 are their masses and r is the distance between them. This was in fact one or Newton’s many great achievements his formulation of this gravitational law. When investigating the gravitational field around a planet change the further an object is from a planet the smaller the force. Therefore the total force surrounding the planet is greater. Field strength is defined as force per unit (1 kg) mass or g = F/m. Size has no effect on the gravitational field and the force exerted on an object because the amount of force stays the same whether the planet is big or small. The field strength is inversely proportional to the square of the distance from the center of the planet. The value of G will always stay constant at 6.67 × 10–11 N m2 kg–2. Each point in a gravitational field has gravitational potential which means that it is work done in bringing unit mass (1 kg) from infinity to that point. Gravitational potential (V) is calculated from the equation V = –Gm1/r, where G is the gravitational constant, m1 is the mass of the planet and r is the distance from the center of the planet. The potential energy of a mass m2 at any point is given by Ep = m2V. Gravitational potential energy increases steadily with height and is directly proportional to height. Increases in gravitational potential energy are directly proportional to the increase in height and mass. The surface of the Earth the gravitational field strength is nearly constant in magnitude, g = 9.81 N kg–1.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-42466515270777931602010-03-22T08:32:54.658-04:002010-03-22T08:32:54.658-04:00Bryan DeVissiere
Hernan Gandarilla
Alex Soares
Sin...Bryan DeVissiere<br />Hernan Gandarilla<br />Alex Soares<br />Sindy Trujillo<br />Period 1<br /><br />During this lesson, we learned about the gravitational field and its components. The gravitation field is the area around an object where another mass will experience a gravitational force. The field is represented by field lines which show the direction of the force on a mass in the field. And thus, simply stated, the closer together the field lines, the stronger the force on the mass. The field strength is the gravitational force per unit (1 kg) mass. This can be written as g = F/m. This field strength at any point is proportional to the mass m1 of the object causing the field, and inversely proportional to the square of the distance r to the point from the centre of the object. This can be written as g = Gm1/r2. The force on a mass m2 in the field is given by F = gm2, which gives F = Gm1m2/r2. Now, the gravitational potential at any point in the field is given by V = –Gm1/r. Throughout this summary we’ve been using the variable G and not know exactly what it means. G is the gravitational constant and its value is 6.67 × 10–11 N m2 kg–2. In relation to little g, the gravitational field strength close to the center of the earth is, in magnitude, g = 9.81 N kg–1. Because the variation in the gravitational field strength is going to be in the negative phase, the potential energy gained by an object moving in the gravitational field can be taken to be directly proportional to the increase in height: (delta)Ep = mg(delta)h. If an object of mass m2 is travelling with speed v in a circular orbit of radius r around a planet of mass m1, then the centripetal force F = m2v2/r is equal to the gravitational force on the object F = Gm1m2/r2.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-25309517662727951782010-03-22T08:25:05.665-04:002010-03-22T08:25:05.665-04:00Talisha Peeler & Sara Atehortua
In this lesson...Talisha Peeler & Sara Atehortua<br />In this lesson we learned that the gravitational field is an area around an object where another mass will experience gravitational force. We also now know that the closer together the fields are the stronger the forces are too. No matter what, magnitude and direction in the field always stay the same. And lastly, the direction of the field is vertically downward at any point on the surface.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-70894776476080879372010-03-08T13:37:05.444-05:002010-03-08T13:37:05.444-05:00I learned that speed is a scalar quantity and only...I learned that speed is a scalar quantity and only has magnitude and not direction, and that Velocity is a vector quantity and has both direction and magnitude. When the engine of the object has reached its maximun spped it had a constant speed and changes in velocity. When and object is moving in a cirlce, the magnitude of the centripetal acceleration relates to that magnitude and velocity because it increases as the velocity increases. The tangenial acceleration vector is 0 when the wheel moves at a constant speed. Speed is magnitude of velocity. a=v2/r is the relashionship b/w centripetal acceleration and speed. Centripetal force= Fmv2/r.<br />In centripetal force the weight and magnitdues are constant. The objects going in a vertical cirlce, the centripetal foce will vary.<br /><br />-Naelia B.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-68765862555510294302010-03-02T09:47:02.797-05:002010-03-02T09:47:02.797-05:00We also learned that the angle of a whole circle i...We also learned that the angle of a whole circle is equal to 2 pi radians. Also, the angular displacement of an object moving in a circle is measured in radians and given the symbol "0". The period of rotation t of an object moving in a circle is given by t=2pi/w. The ideas and equations of circular motion can be applied to the motion of objects on all scales, including subatomic particles.<br />Angelo<br />Dondre<br />DamienAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-41992897947147402852010-03-01T09:41:50.344-05:002010-03-01T09:41:50.344-05:00We learned that an object moving in a circle alway...We learned that an object moving in a circle always have a changing velocity directed at a tangent to the circle and a changing centripetal acceleration directed towards the center of the circle. For uniform circular motion there is no tangential acceleration, a centripetal acceleration which is constant in magnitude, changing direction and directed to the center of the circle and a centripetal force which is constant in magnitude, changing direction and directed to the center of the circle. The SI unit of the angle is called the radian. The equation for centripetal acceleratiion is a=v^2/r where v is the speed of the object and r is the radius of the circle.<br />Dondre <br />Damien<br />AngeloAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-3165683358408997892010-03-01T09:11:06.065-05:002010-03-01T09:11:06.065-05:00We found out that an object that moves in a circle...We found out that an object that moves in a circle has a changing velocity tangent to the circle and a changing centripetal acceleration towards the center of the circle. Objects moving in a circle at constant speeds have a uniform circular motion which has no tangential acceleration, a velocity which is constant and tangent to the circle, and centripetal acceleration changing direction directed towards the center of the circle. The formula for centripetal acceleration is a=v2/r and the equation for centripetal force is F=mv2/r.<br /> The SI unit of an angle is called the radian. The radian is a unit of plane angle, equal to 180/π (or 360/(2π)) degrees, or about 57.2. The angle of a full circle is equal to 2π, by the same principle; a quarter of a circle is equal to π/2, and half a circle is π. The length of an arc in a circle is equal to r (angle), where the angle is in radians and r is the radius of the circle. As an object moves in a circular motion, its angular displacement can be described in radians i.e. half a circle is described as π and not 180 degrees. Centripetal acceleration is the rate of change of tangential velocity, the direction of the centripetal acceleration is always inwards along the radius vector of the circular motion. Centripetal acceleration is equal to r(w)to the second power. Within the circle, velocity can be measured as radians per second, just as they are used to measure displacement. Centripetal force can be written as mr(w)to the second power.<br /> We learned that circular motion is able to be applied to all forms of objects known to man, the principles and equations can be used on all types or sizes of objects. In addition, centripetal forces can be in different types of forms, for example path of a roller coaster or the path of a race car track. The centripetal force must be provided by friction alone on a curve, and a increase in speed could lead to an unexpected skid if friction is insufficient can create something very unpleasant if there is no friction.<br /><br />Boronny Touch<br />Adrian Simpson<br />Sergio MunozAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-81385306635080254042010-03-01T09:03:51.229-05:002010-03-01T09:03:51.229-05:00Cynthia Garcia
Precious Humphrey
Elizabeth Trejo
R...Cynthia Garcia<br />Precious Humphrey<br />Elizabeth Trejo<br />Roneisha Handy<br />Selina Meach<br />Phyllisse Lewis<br /><br />Objects in circular motion have changing velocities because they’re changing their directions of motion. Uniform circular motion is when an object moves in a circle at a constant speed. The equation for centripetal acceleration is velocity squared divided by the radius. The equation for centripetal force is mass times velocity squared divided by the radius. 2π radians are equal to the angle of a whole circle since 2π equals 360º. We use radians to measure the angular displacement of an object moving in a circle.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-42962633794713044502010-02-24T08:58:29.678-05:002010-02-24T08:58:29.678-05:00We learned that and object in a circle will always...We learned that and object in a circle will always have a changeing velocity directed at a tangent to the circle.<br />The equation for centripetal acceleration is a=v^2/r, where v is the spped of the object and r is the radius of the circle.<br />The equation for the centripetal force is F=mv^2/r, where m is the mass of the object moving in a circle. The angle of a whole circle is 2pi radians. The period of rotation of an object moving in a circle is given by t=2pi/w. the centripetal acceleration can be written as a=rw^2. The centripetal force can be written as F=mrw^2. The angular velocity of an object moving in a circle is measured in radians per second and given the symbol w. <br /><br />By Kevin E. Devissiere<br />Victoria Taiwo<br />Margarita RodriguezAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-64217582154263885322010-02-12T08:44:15.069-05:002010-02-12T08:44:15.069-05:00Nefertiti Ferguson-Goforth; Marissa Martinez, Lizz...Nefertiti Ferguson-Goforth; Marissa Martinez, Lizzette Martinez <br />You don’t need an arrow for speed because speed is a scalar quantity and only has magnitude, not direction.<br />You need an arrow for velocity because Velocity is a vector quantity and has direction and magnitude.<br />The speed of an engine is constant when the engine has reached it maximum speed. Velocity is never constant when the engine reaches it maximum speed.<br />The wheel only has tangential acceleration when the wheel is speeding up and slowing down.<br />The tangential acceleration and velocity are the same when an object, such as a Ferris wheel, speeds up but when the object slows down the tangential acceleration and the velocity are inverse. <br />Centripetal acceleration increases as the magnitude of velocity increases.<br />When an object reaches constant speed, its tangential acceleration is zero.<br />Formula: a = v2/rAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-34349972138658650962010-02-12T08:44:03.377-05:002010-02-12T08:44:03.377-05:00Within this section of physics, we learn about cir...Within this section of physics, we learn about circular motion and its effects on how everything moves around us. We learned that an object moving in a circle will always have a changing velocity directed at a tangent to the circle and along with this, we learned that a changing centripetal acceleration directed towards the center of the circle. Centripetal acceleration is the rate of change of tangential velocity. Now, an object moving in a circle at constant speed is said to have uniform circular motion. The rules for uniform circular motion there are that there are no tangential acceleration, there must be a velocity which is constant in magnitude, there must a changing direction and directed at a tangent to the circle, there must be a centripetal acceleration which is constant in magnitude, changing direction and directed to the center of the circle, and also a centripetal force which is constant in magnitude, changing direction and directed to the center of the circle.<br /> The equation for centripetal acceleration is:<br /> a = v2/r, where v is the speed of the object and r is the radius of the circle. <br />The equation for centripetal force is: <br />F = mv2/r, where m is the mass of the object moving in a circle.<br /><br />bryan,hernan, Mohammed, Sindy.....We are not finishedAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-53114526967998126402010-02-12T08:43:38.732-05:002010-02-12T08:43:38.732-05:00Alyssa Zargos, Alex Soares, Kevin Andrade.
Speed ...Alyssa Zargos, Alex Soares, Kevin Andrade.<br /><br />Speed and velocity in a moving object going in a circle differs. The speed of an object moving in a circle only has magnitude, and no direction. Also velocity has direction and magnitude. During moving objects, the speed is constant when it has reached maximum speed, and it changes as it starts up and stops. Also the velocity is never constant because it is always slowing down and starting up. An object moving at a constant speed in a circle is said to have uniform circular motion. When circular motion is uniformed tangential acceleration is equal to zero. An object moving in a circle would always a changing velocity at a tangent to the circle. Also, there is a changing centripetal acceleration directed towards the centre of the circle and perpendicular to the velocity. The equation a = v2/r is to figure out centripetal acceleration, and to figure out centripetal force is F = mv2/r. The constants v = speed of the object, r = radius of the circle, and m = the mass of the object moving. <br /><br />There are a few different ways to measure angles. The SI unit of the angle is called a radian. The radian is defined as the angle that starts at the center of the circle by an arc on the circumference that is equal in length to the radius of the circle. The angle of a whole circle is 2pi radians. The length of any arc is equaled to r-theta. Theta equals the angle in radians. The angular displacement of an object moving in a circle is measured in radians and given the symbol theta. The centripetal acceleration is known as a = rw, and the force is equaled as F = mrw2. <br /><br />The principles and equations of circular motion can be applied to the motion of objects on all scales, including subatomic particles. The centripetal force can be provided by a variety of different forces and combinations of forces. The contribution of weight of an object to the centripetal force will vary.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-8070608540488482462010-02-12T08:43:37.795-05:002010-02-12T08:43:37.795-05:00Sherwin Yu
Emmanuel Torres
Kevin Pham
Circles ar...Sherwin Yu<br />Emmanuel Torres<br />Kevin Pham<br /><br /> Circles are very critical in physics for there are many different types of forces acting upon an object traveling in that circle. When an object is traveling in a circle the direction is constantly changing, therefore the velocity will always be changing. The speed is only constant whenever it has reached maximum speed. Centripetal force and acceleration are always directed towards the center of the circle. An object moving in a circle will always have a centripetal acceleration that is perpendicular to the velocity.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-84431147573157602362009-05-18T09:16:00.000-04:002009-05-18T09:16:00.000-04:00Niccolai Arenas, Yesenia Llanos, Catalina Estrada,...Niccolai Arenas, Yesenia Llanos, Catalina Estrada, Nataliya Kushta said...<br />Yesenia Llanos <br />Catalina Estrada<br />Nataliya Kushta<br />Niccolai Arenas<br />May 12, 2009<br />Multimedia Lab<br /><br />Oscillations<br /><br />-Why does the length of the acceleration indicator bar fall back to zero after the cannonball leaves the cannon?<br />Because the cannon ball is not moving anymore.<br />-Describe the relationship between force and displacement for the cannonball as it travels along the barrel.<br />The force is independent of the displacement.<br />-How is the graph is consistent with the relationship F = ma?<br />The graph is a straight line that passes through the origin, showing that force is proportional to acceleration.<br />-What is the displacement when force is zero<br />Zero<br />-Describe the relationship between force and displacement<br />The force is proportional to the displacement but they are going in opposite directions.<br />-How does this new oscillation affect the relationship between force and displacement?<br />They are the same<br />-What is the displacement when acceleration is zero?<br />It is zero<br />What do you notice about the relative directions of force and acceleration?<br />They are always in the same direction<br />-Describe the relationship between force and acceleration.<br />Force is proportional to the acceleration<br />-What is the acceleration when displacement is a) zero, and b) at its maximum value?<br />a. Zero<br />b. Will also be its maximum value<br />-What are the relative directions of displacement and acceleration?<br />Opposite directions<br />-What has happened to the gradient of the graph?<br />Increases<br />-a.) Which is constant while the ball is in the barrel and then falls to zero after the ball has left?<br />Acceleration<br />-b) which increases steadily while the ball is in the barrel?<br />Velocity<br />-What are the main differences between the nature of the force on the cannonball and the nature of force that produces oscillation?<br />The force on the cannonball is constant and in the same direction as the displacement<br />-Are displacement and velocity always in opposite directions?<br />No<br /><br />-What is the displacement when velocity has its maximum value?<br />Zero<br />-Where is the boxing glove when it exhibits its fastest movement?<br />At the equilibrium position<br />-What do you notice about the value of acceleration when velocity is zero?<br />It’s at its maximum<br />-What is the velocity when displacement has its maximum values?<br />Zero<br />-What is acceleration when displacement is zero?<br />Zero<br />-Of the variables frequency, period, angle and angular velocity, which are time-dependent and which are constant?<br />Frequency, period and angular velocity remain constant. And the angle is time-dependent.<br />-What do you notice about the values of 2πft angle θ?<br />They are the same<br />-What range of values may sin 2πft have?<br />It is between -1 and 1<br />-What range of values may x have?<br />Has an interval of .02<br />-What is the only difference between the motions described by the equation x = A sin 2πft and the equation x = A cos 2πft?<br />The first equation has an initial displacement of zero and the second has the initial displacement A.<br /><br />Wave Behavior<br /><br />-Which way are the reflected waves moving?<br />It goes downwards<br />-How would you make the ripple tank produce waves of a smaller wavelength?<br />Make the motor run faster so that the wooden beam vibrates more frequently.<br />-What happens to the plane wave after reflection?<br />It curves<br />-What happens to the plane wave after reflection?<br />They get smaller as they get to the center.<br />-What shape does the reflected wave have?<br />A circular shape.<br />- Where do the reflected waves seem to come from?<br />They seem to come from a point behind the reflector that is the same distance from it as the original source of the waves<br />-What happens to the wave?<br />It changes direction when it enters shallow water<br />- What happens to the speed of the waves?<br />The wave’s speed is less in the shallow water<br /><br />-What happens to the wavelength?<br />The wavelength is reduced in the shallow water<br />- Which way is the wave moving?<br />From left to right<br />- Focus on any particle. Which way are the particles moving?<br />Up and down<br />- Assuming the background lines are 1 cm apart, how high is your wave above the black centre line?<br />1cm<br />- What is the amplitude now?<br />7.5 cm<br />-Assuming the background lines are approximately 1 cm apart, what is the distance between two wave tops?<br />15 cm<br />-What is the amplitude of the wave?<br />5 cm<br />- Move the Wavelength slider to the far left. What happens?<br />The wave crest gets close together<br />- What is the wavelength now?<br />10 cm<br />- What is the wavelength now?<br />20 cm<br />- What is the amplitude now?<br />5 cm<br />- Have you changed the energy carried by the wave? Explain your answer<br />No, because the wave amplitude does not change<br />- What is the amplitude of the wave and what is its wavelength?<br />Its amplitude is 5 cm and its wavelength is 15 cm.<br />- What has happened to the velocity of the wave?<br />It has increased<br />- Click and measure the amplitude of the wave and wavelength again. Have they changed?<br />No<br />May 12, 2009 9:02 AM <br />Pauline Santos, Kevin Ith, Yoshika Wason, Hazel Mabesa said...<br />1) Why does the length of the acceleration indicator bar fall back to zero after the cannonball leaves the cannon?<br />The cannonball is no longer experiencing an accelerating force. Acceleration is zero and velocity (neglecting air resistance) is constant.<br /><br />2) Describe the relationship between force and displacement for the cannonball as it travels along the barrel<br />The force is independent of displacement<br /><br />3) How is the graph is consistent with the relationship F = ma?<br />The graph is a straight line that passes through the origin, showing that force is proportional to acceleration. The relationship F = ma applies to all motion of any constant mass, and it shows that the gradient of the graph is the mass, m.<br /><br />4) What is the displacement when force is zero?<br />Zero<br /><br />5) Describe the relationship between force and displacement<br />Force is proportional to displacement, but in the opposite direction.<br /><br />6) How does this new oscillation affect the relationship between force and displacement?<br />The relationship is the same as before (but the maximum values of force and displacement are smaller).<br /><br />7) What is the displacement when acceleration is zero?<br />It is also zero.<br /><br />8) What do you notice about the relative directions of force and acceleration?<br />They are always in the same direction.<br /><br />9) Describe the relationship between force and acceleration.<br />Force is proportional to acceleration<br /><br />10) What is the acceleration when displacement is a) zero, and b) at its maximum value?<br />a) Zero<br />b) At its maximum value<br /><br />11) What are the relative directions of displacement and acceleration?<br />They are always in the opposite direction.<br /><br />12)What has happened to the gradient of the graph?<br />It has increased.<br /><br />13)Of the variables displacement, velocity and acceleration (and ignoring the effects of air resistance):<br />a) which is constant while the ball is in the barrel and then falls to zero after the ball has left? Acceleration<br />b) which increases steadily while the ball is in the barrel? Velocity <br /><br />14) What are the main differences between the nature of the force on the cannonball and the nature of force that produces oscillation? <br />The force on the cannonball is constant and in the same direction as the displacement. A force producing oscillation varies with displacement and is in the opposite direction to it, thus acting as a restoring force. <br /><br />15)Looking at the indicator bars, are displacement and velocity always in opposite directions? <br />No <br /><br />16)What is the displacement when velocity has its maximum value?<br />Zero<br /><br />17) Where is the boxing glove when it exhibits its fastest movement?<br />At the equilibrium position.<br /><br />18) What do you notice about the value of acceleration when velocity is zero?<br />It is at its maximum.<br /><br />19) What is the velocity when displacement has its maximum values?<br />Zero<br /><br />20) What is acceleration when displacement is zero?<br />Zero<br /><br />21) Of the variables frequency, period, angle and angular velocity, which are time-dependent and which are constant? <br />Angle is time-dependent. Frequency, period and angular velocity remain constant.<br /><br />22) What do you notice about the values of 2πft angle θ?<br />They are the same.<br /><br />23) What range of values may sin 2πft have? <br />Between –1 and +1.<br /><br />24) What range of values may x have? <br />Between –A and +A.<br /><br />25) What is the only difference between the motions described by the equation x = A sin 2πft and the equation x = A cos 2πft?<br />The first equation has an initial displacement of zero and the second has the initial displacement A.<br />May 12, 2009 9:50 AM <br />Jon Terrence Tailane said...<br />1. The cannonball is no longer experiencing an accelerating force because its slowing down.<br /><br />2. The force is independent of displacement.<br /><br />3. The graph is a straight line that passes through zero. F = ma applies to all motion of any constant mass.<br /><br />4. When force is zero, so is displacement<br /><br />5. Force is proportional to displacement, but in the opposite direction.<br /><br />6. The relationship is the same as before but the maximum values of force and displacement are smaller.<br /><br />7. When acceleration is zero, so is displacement<br /><br />8. Relative direction of force and acceleration are relatively the same.<br /><br />9. Force is proportional to acceleration.<br /><br />10. The relative direction of acceleration and displacement are always opposite<br /><br />11. The force on the cannonball is constant and in the same direction as the displacement. A force producing oscillation varies with displacement and is in the opposite direction to it, thus acting as a restoring force.<br /><br />12. Displacement and velocity aren’t always opposite<br /><br />13. Displacement is zero when velocity is at its maximum value<br /><br />14. The boxing glove exhibits the fastest speed at its equilibrium <br /><br />15. Acceleration is at maximum value when velocity is zero<br /><br />16. Angle is time-dependent. Frequency, period and angular velocity are constant<br /><br />17. Both values of 2πft and angle θ are the same<br /><br />18. between -1 & 1<br /><br />19. Between –A and +A.<br /><br />20. First equation has an initial displacement of zero and the second has the initial displacement of A.<br />May 12, 2009 10:01 AM <br />David, Ngoc, Victoria said...<br />David Kristy, Ngoc Nguyen, Victoria Nguyen<br /><br />Why does the length of the acceleration indicator bar fall back to zero after the cannonball leaves the cannon?<br /><br />The cannonball is no longer experiencing an accelerating force. Acceleration is zero and velocity (neglecting air resistance) is constant.<br /><br />Describe the relationship between force and displacement for the cannonball as it travels along the barrel.<br />The force is independent of displacement<br /><br />How is the graph is consistent with the relationship F = ma?<br />The graph is a straight line that passes through the origin, showing that force is proportional to acceleration. The relationship F = ma applies to all motion of any constant mass, and it shows that the gradient of the graph is the mass, m.<br /><br />What is the displacement when force is zero?<br />Zero<br /><br />Describe the relationship between force and displacement <br />Force is proportional to displacement, but in the opposite direction. <br /><br />How does this new oscillation affect the relationship between force and displacement? <br />The relationship is the same as before (but the maximum values of force and displacement are smaller). <br /><br />What is the displacement when acceleration is zero? <br />It is also zero. <br /><br />What do you notice about the relative directions of force and acceleration? <br />They are always in the same direction. <br /><br />Describe the relationship between force and acceleration <br />Force is proportional to acceleration. <br /><br />What is the acceleration when displacement is a) zero, and b) at its maximum value? <br />a) Zero<br />b) At its maximum value<br /><br />What are the relative directions of displacement and acceleration? <br />They are always in the opposite direction. <br /><br />What has happened to the gradient of the graph? <br />It has increased. <br /><br /><br />Of the variables displacement, velocity and acceleration (and ignoring the effects of air resistance): <br />which is constant while the ball is in the barrel and then falls to zero after the ball has left? <br />Acceleration <br /><br />which increases steadily while the ball is in the barrel? <br />Velocity <br /><br />What are the main differences between the nature of the force on the cannonball and the nature of force that produces oscillation? <br />The force on the cannonball is constant and in the same direction as the displacement. A force producing oscillation varies with displacement and is in the opposite direction to it, thus acting as a restoring force. <br /><br />Looking at the indicator bars, are displacement and velocity always in opposite directions? <br />No <br /><br />What is the displacement when velocity has its maximum value?<br />Zero<br />May 13, 2009 9:02 AM <br />Rai'jona Crear, Carlos Velazquez, Cyril Senu said...<br />Raijona Crear<br />Carlos Velázquez<br />Cyril SENU <br /><br />Oscillations<br /><br />Why does the length of the acceleration indicator bar fall back to zero after the cannonball leaves the cannon?<br />The cannonball is no longer experiencing an accelerating force. Acceleration is zero and velocity (neglecting air resistance) is constant.<br /><br />Describe the relationship between force and displacement for the cannonball as it travels along the barrel. <br />The force is independent of displacement.<br /><br />How is the graph consistent with the relationship F = ma?<br />The graph is a straight line that passes through the origin, showing that force is proportional to acceleration. The relationship F = ma applies to all motion of any constant mass, and it shows that the gradient of the graph is the mass, m. <br /><br />What is the displacement when force is zero? <br />Zero.<br /><br />Describe the relationship between force and displacement. <br />Force is proportional to displacement, but in the opposite direction. <br /><br />How does this new oscillation affect the relationship between force and displacement? <br />The relationship is the same as before (but the maximum values of force and displacement are smaller). <br /><br />What is the displacement when acceleration is zero? <br />It is also zero.<br /><br />What do you notice about the relative directions of force and acceleration?<br />They are always in the same direction.<br /><br />Describe the relationship between force and acceleration.<br />Force is proportional to acceleration.<br /><br />What is the acceleration when displacement is a) zero, and b) at its maximum value?<br />A) Zero<br />B) At its maximum value<br /><br />What are the relative directions of displacement and acceleration?<br />They are always in the opposite direction.<br /><br />A) Which is constant while the ball is in the barrel and then falls to zero after the ball has left? Acceleration<br />B) Which increases steadily while the ball is in the barrel? Velocity<br />What are the main differences between the nature of the force on the cannonball and the nature of force that produces oscillation?<br />The force on the cannonball is constant and in the same direction as the displacement. A force producing oscillation varies with displacement and is in the opposite direction to it, thus acting as a restoring force.<br />Looking at the indicator bars, are displacement and velocity always in opposite directions? NO<br />What is the displacement when velocity has its maximum value? Zero<br />Where is the boxing glove when it exhibits its fastest movement?<br />At the equilibrium position.<br />What do you notice about the value of acceleration when velocity is zero?<br />It is at its maximum.<br />What is the velocity when displacement has its maximum values?<br />Zero<br />What is acceleration when displacement is zero?<br />Zero<br />Of the variables frequency, period, angle and angular velocity, which are time-dependent and which are constant?<br />Angle is time-dependent. Frequency, period and angular velocity remain constant. <br />What do you notice about the values of 2πft angle θ?<br />They are the same.<br />What range of values may sin 2πft have?<br />Between –1 and +1.<br /><br />What range of values may x have?<br />Between –A and +A.<br /><br />What is the only difference between the motions described by the equation x = A sin 2πft and the equation x = A cos 2πft?<br />The first equation has an initial displacement of zero and the second has the initial displacement A.<br /><br />Which way are the reflected waves moving?<br />Downwards, across the original wave.<br /><br />How would you make the ripple tank produce waves of a smaller wavelength?<br />Make the motor run faster so that the wooden beam vibrates more frequently.<br /><br />What happens to the plane wave after reflection?<br />It becomes a circular wave.<br /><br />What happens to the curved reflected waves?<br />They get smaller and concentrate into a point.<br /><br />What shape does the reflected wave have?<br />A circular shape.<br /><br />Where do the reflected waves seem to come from?<br />They seem to come from a point behind the reflector that is the same distance from it as the original source of the waves.<br /><br />What happens to the wave?<br />The wave changes direction when it enters the shallow water.<br /><br />What happens to the speed of the waves?<br />The wave speed is less in the shallow water.<br /><br />What happens to the wavelength?<br />The wavelength is reduced in the shallow water.<br /><br />Which way is the wave moving?<br />It is moving from left to right.<br /><br />Which way are the particles moving?<br />Up and down.<br /><br />How high is your wave above the center black line?<br />One cm.<br />What is the amplitude now?<br />7.5 cm<br /><br />What is the distance between two wave tops?<br />15 cm.<br /><br />What is the amplitude of the wave?<br />5 cm.<br /><br />What happens when you move the wavelength slider to the left ?<br />The wave crests get closer together<br /><br />What is the wavelength now?<br />10 cm.<br /><br />What is the wavelength now?<br />20 cm<br /><br />What is the amplitude now?<br />5cm<br /><br />Have you changed the energy carried by the wave?<br />No, because the wave amplitude has not changed.<br /><br />What is the amplitude of the wave and what is its wavelength?<br />Its amplitude is 5 cm and its wavelength is 15 cm.<br />May 13, 2009 9:05 AM <br />Olivia Thergood, Daniel Maloney, Katiria Lopez said...<br />Olivia Thergood<br />Daniel Maloney <br />Katiria Lopez<br />Physics Per. 1 <br />Sound Waves <br /><br />1. Wavelength = velocity / frequency <br />= 340 m/s / 1000 Hz <br />= .34 <br />2. Cancellation would not occur with stereo sound.<br />May 14, 2009 9:59 AM <br />Paige, Beth, Erica said...<br />Read through the page, answer any questions that are in the "lecture" part and summarize the content. Turn in the answers and summary next class.<br /><br />Summary<br /><br />A stone dropped in water causes a disturbance to travel outward in an expanding circle. The water doesn't go anywhere; it is only the energy which moves.<br /><br />V= (gd)^1/2<br /><br />If a succession of stones were dropped, one each second, a wave train would be created. The expanding circles in the wave train are called wave fronts. <br />1.) One stone is dropped into the water every 1/5 second. What are the period and frequency of the wave motion?<br />Period = 1/5 second<br />Frequency = 1 / (1/5) = 5 Hz<br />The distance between the maxima is called the wavelength of the wave.<br /><br />Distance = speed x time<br />Wavelength = wave speed x period<br /><br />Assuming the wave speed of the disturbance on water is 2 m/s, and the period of the wave motion is 1/5 s, what is the wavelength of the wave motion?<br />Wavelength = (2 m/s) x (1/5 s) = 0.4 m<br />Longitudinal: displacements are parallel to direction of propagation. Transverse: displacements perpendicular to propagation direction Compressions are regions of above-normal air pressure. Rarefactions are regions of below-normal air pressure.<br />2.) The speed of sound in air is about 340 m/s. What is the wavelengh of sound created at 1000 Hz?<br />Waverlength = (340) x (1000) = 340000<br />Ultrasound is sound at a frequency which is outside of the range of human hearing. Sound travels faster in warm air. When compressions from one source overlap rarefactions from another, cancellation occurs. Beat frequency<br />is the difference between the frequencies of the two tuning forks.<br />Positions of zero rope displacement in a standing wave are called nodes. At each end of any loop there is a node, and two loops makes one wavelength. Distance between adjacent nodes is one-half wavelength. <br />Doppler Effect - Approaching, the frequency of a sound is higher because the wavefronts are closer together<br />in time. Departing, the frequency is lower. Sounds are a mixture of the fundamental and one or more of the overtones. These mixtures are called composite vibrations.<br />May 14, 2009 10:01 AM <br />Yesenia Llanos, Niccolai Arenas, Ashley Lynk, Catalina Estrada, Nataliya Kushta said...<br />• Move the amplitude slider back and forth to see its effects on all parts of the wave. What happens?<br />The amplitude changes the height of the crests and the depth of the troughs, but not the wavelength or velocity.<br />• Move the Wavelength slider back and forth to see its effect on all parts of the wave. What happens?<br />The wavelength changes the distance between neighboring crests and troughs, but not the amplitude or velocity.<br />• Where is the other red particle at this moment?<br />There at the same height.<br />• Where is the yellow particle at this moment?<br />At the lowest point.<br />• Where is the other red particle at this moment?<br />At the lowest point.<br />• Where is the yellow particle at this moment?<br />At its highest point.<br />• How far apart are the two red particles?<br />One wavelength.<br />• How far apart are the red and yellow particles?<br />Half a wavelength.<br />• Which ones get through both of the circular filters?<br />The vertical ones<br />• What happens to the vertical waves now?<br />They don’t pass through the filter<br />• What happens to the horizontal waves now?<br />They get through the first filter but don’t go through the second one.<br />• How could you allow the horizontal waves to get through the second filter?<br />By rotating it 90 degrees.<br />• Look at the length of the wave on the scale. How far did the wave travel?<br />24 m<br />• What is the connection between the numbers in each row of the table?<br />velocity = frequency X wavelength<br />• Which way is the wave moving?<br />Left to right<br />• Look at the particles in the material. How is their disturbance different to the way they are disturbed by a transverse wave?<br />Here the particles go from left to right.<br />• Assuming the background lines are 1 cm apart, what is the maximum displacement of a particle from the middle of its motion?<br />0.5 cm<br />• What is the amplitude now?<br />0.25 cm.<br />• How did you do this?<br />By increasing the amplitude.<br /><br />• Assuming the vertical background lines are 1 cm apart, what is the distance between the centres of two neighbouring compressions?<br />10 cm.<br />• What is the distance between the centres of two neighbouring rarefactions?<br />10 cm<br />• How does this compare to the distance between compressions?<br />It is the same.<br />• What is the wavelength now?<br />About 13 cm<br />• What is the amplitude of the wave?<br />0.5 cm<br />• What is the wavelength of the wave?<br />10 cm<br />• What has happened to the velocity of the wave?<br />Increases<br />• What is the amplitude of the wave now?<br />0.5 cm<br />• What is the wavelength of the wave now?<br />10 cm<br />• What do you notice about the amplitude and wavelength of the wave as the velocity changes? <br />Amplitude and wavelength stay the same.<br />• Using this information, why do you think you see a lightning flash before hearing thunder?<br />Light travels faster than sound.<br />• Find the minimum wavelength value.<br />5 cm<br />• What is the maximum wavelength value?<br />20 cm<br />• How do they move?<br />Back and forth<br />• As a result of the passage of the wave, how have the layers either side of this one been moved from their rest position on one of the vertical lines?<br />They move closer to the layer.<br />• As a result of the passage of the wave, how have the layers either side of this one been moved from their rest position on one of the vertical lines?<br />They moved further from the layer.<br />• From the graph, what is the approximate displacement of the green layers at the middle of a compression?<br />Zero<br />• From the graph, what is the approximate displacement of the red layers at the middle of a rarefaction?<br />Zero<br /><br /><br />• When the displacement graph is positive, which way have the particles been displaced from their rest positions?<br />To the right.<br />• When the displacement graph is negative, which way have the particles been displaced from their rest positions?<br />To the left.<br />• How does this affect the air pressure?<br />Decreases <br />• What is the height in divisions of the crest of the wave on the screen?<br />Two<br />• What is the amplitude of the wave, in volts?<br />2 × 0.2 = 0.4 V<br />• What is the wavelength, in divisions, of the wave on the screen?<br />Four <br />• What time is represented by the wavelength value?<br />4 × 0.5 = 2 ms = 2/1000 in seconds<br />• Use this equation to calculate the frequency of the sound you see being displayed on the oscilloscope screen.<br />1000/2 = 500 Hz<br />• What is the periodic time and the frequency of this sound?<br />About 12 ms and 80 Hz.<br /><br /><br />Combining Waves<br /><br />• How do their amplitudes and wavelengths compare?<br />They are the same.<br /><br />• Are the waves in phase or out of phase with each other?<br />They are in phase.<br /><br />• How do these two waves compare in amplitude and wavelength?<br />They are the same.<br /><br />• Predict what will happen when the two waves are added.<br />They cancel out<br /><br />• How do the two waves compare in amplitude?<br />The amplitude of the blue wave is one third of the amplitude of the red wave.<br /><br />• How do they compare in wavelength?<br />The wavelength of the blue wave is one third of the wavelength of the red wave.<br /><br />• How do they compare in frequency?<br />The frequency of the blue wave is three times the frequency of the red wave.<br />• How many times does this resonance happen?<br />Four<br /><br />• What are these frequencies?<br />15, 30, 45, and 60 units.<br /><br />• How many antinodes are seen at each of these frequencies?<br />One, two, three, and four antinodes, respectively.<br /><br />• What is the ratio of the frequencies that cause resonance?<br />1:3:5<br /><br />• So, from the length of your first resonant position, what is the wavelength of the sound you have heard?<br />4 × 0.215 = 0.860 m<br /><br />• From this equation and your wavelength value, what is the velocity of the sound in the air of the tube for this frequency of tuning fork?<br />384 × 0.860 = 330 m s–1<br /><br />• How does this length compare with the first one?<br />It is three times as long.<br /><br />• Repeat the experiment and see how many resonant positions you can find. How many are there?<br />Three<br /><br />• From the length of the second one, where there are three node-antinode distances, what is the wavelength and velocity of the sound?<br />0.688 m and 330 m s–1.<br /><br />• What happens to the wave?<br />The wave spreads around the edges of the two obstacles.<br /><br />• What difference do you see in the waves that diffract through the narrow gap?<br />The waves are diffracted through a wider angle when they pass through a narrow gap.<br /><br />• What do you notice about the area to the right of the barrier compared to the single-slit picture?<br />There are two regions where the wave crests are missing.<br /><br />• What would you expect to happen where two wave crests meet?<br />A very high crest would be formed.<br /><br />• What would you expect to happen where two wave troughs meet?<br />A very low trough would be formed.<br /><br />• Now look for places where a crest from one wave source meets a trough from the other. What would you expect to happen in these places?<br />The crest of one wave is cancelled out by the trough of the other one, leaving the water calm.<br /><br />• What happens at the point marked with a red dot?<br />A wave crest from source 1 has met a crest from source 2 so they reinforce and make a high crest.<br /><br />• What is the difference between your two answers in terms of wavelengths?<br />A wave crest from source 1 has met a crest from source 2 so they reinforce and make a high crest.<br /><br />• How do these differ from the other cases?<br />They are both cancellations, where a crest from one source meets a trough from the other.<br /><br />• What do these straight lines represent?<br />Regions of cancellation where crests from one source meet troughs from the other.<br /><br />• What does this do to x?<br />Increasing the source separation a decreases x.<br /><br />• What does this do to x?<br />Increasing the wavelength increases x.<br /><br />• What does this do to x?<br />Increasing the distance to the cross-section increases x.<br />May 14, 2009 10:32 AM <br />Lucas, Vince, Kelvine, Oscar, Abdou said...<br />Vince Nguyen<br />Oscar Osorio <br />Lucas Leite<br />Abdou Williams<br />Kelvine Clarke<br /><br />Wave Behavior<br /><br />Which way are the reflected waves moving? <br />- Downwards, across the original wave.<br /><br />How would you make the ripple tank produce waves of a smaller wavelength?<br />- make the motor run faster so that the wooden beam vibrates more frequently.<br /><br />What happens to the plane wave after reflection?<br />- it becomes a circular wave.<br /><br />What happens to the curved reflected waves?<br />- They get smaller and concentrate into a point.<br /><br />What shape does the reflected wave have?<br />- a circular shape.<br /><br />Where doe the reflected waves seem to come from?<br />- They seem to come from a point behind the reflector that is the same distance from it as the original source of the waves.<br /><br />What happens to the wave?<br />- the wave changes direction when it enters the shallow water.<br /><br />What happens to the speed of the waves?<br />- the wave speed is less in the shallow water.<br /><br />What happens to the wavelength?<br />- the wavelength is reduced in the shallow water.<br /><br />Which way is the wave moving?<br />- from left to right.<br /><br />Focus on any particle. Which way are the particles moving?<br />- from up to down.<br /><br />Assuming the background lines are 1 cm apart, how high is your wave above the black centre line?<br />- 1cm<br /><br />What is the amplitude now? <br />- 7.5 cm<br /><br />What is the amplitude of the wave?<br />- 5 cm<br /><br />Assuming the background lines are approximately 1 cm apart, what is the distance between two wave tops?<br />- 15 cm<br /><br />Move the wavelength slider to the left. What happens?<br />- the wave crests get closer together.<br /><br />What is the wavelength now?<br />- 10 cm<br /><br />What is the wavelength now?<br />- 20 cm<br /><br />What is the amplitude now?<br />- 5 cm<br /><br />Have you changed the energy carried by the wave? Explain your answer.<br />- no, because the wave amplitude has not changed.<br /><br />What is the amplitude of the wave and what is its wavelength?<br />- Its amplitude is 5 cm and its wavelength is 15 cm.<br /><br />What has happened to the velocity of the wave?<br />- it has increased.<br /><br />Click pause and measure the amplitude of the wave and wavelength agaian. Have they changed? <br />- no<br /><br />Move the Amplitude slider back and forth to see its effect on all parts of the wave. What happens?<br />- The amplitude changes the height of the crests and the depth of the troughs, but not the wavelength or velocity.<br /><br />Move the wavelength slider back and forth to see its effects on all parts of the wave. What happens?<br />- the wavelength changes the distance between neighboring crests and neighboring troughs, but not the amplitude or velocity.<br /><br />Where is the other red particle at this moment?<br />- at the same height.<br /><br />Where is the yellow particale at this moment?<br />- at its lowest point.<br /><br />Where is the other red particale at this moment?<br />- at its lowest point.<br /><br />Where is the yellow particale at this moment?<br />- At its highest point.<br /><br />How far apart are the two red particales?<br />- one wavelength.<br /><br />How far apart are the red and yellow particales?<br />- half a wavelength.<br /><br />Which ones get through both of the circular filters?<br />- the vertical ones.<br /><br />What happens to the vertical waves now?<br />- they cannot pass through the filter.<br /><br />What happens to the horizontal waves now? <br />- they get through the first filter but not the second one.<br /><br />How could you allow the horizontal waves to get through the second filter?<br />- by rotating it 90 degrees.<br /><br />Look at the length of the wave on the scale. How far did the wave travel?<br />- 24 m<br /><br />What is the connection between the numbers in each row of the table?<br />- velocity = frequency x wavelength.<br /><br />Which way is the wave moving?<br />- from left to right.<br /><br />Look at the particales in the material. How is their disturbance different to the way they are disturbed by a transverse wave?<br />- Here the particles are disturbed from left to right, rather than up to down.<br /><br />Assuming the background lines are 1 cm apart, what is the maximum displacement of a particle from the middle of its motion?<br />- 0.5 cm<br /><br />What is the amplitude now? <br />- about 0.25 cm<br /><br />How did you do this?<br />- by increasing the amplitude.<br /><br />Assuming the vertical background lines are 1 cm apart, what is the distance between the centres of two neighboring compressions? (particles at the center of each compression should be on, or close to a vertical line.)<br />- approximately 10 cm.<br /><br />what is the distance between the centers of two neighboring rarefactions?<br />( again, particles at the center of each rarefactions should be on, or close to a vertical line.)<br />- approximately 10 cm.<br /><br />How does this compare to the distance between compression?<br />- it is the same.<br /><br />Click pause to pause the wave. What is the wavelength now?<br />- about 13 cm.<br /><br />what is the wavelength of the wave?<br />- 0.5 cm<br /><br />What is the wavelength of the wave?<br />- 10 cm<br /><br />What has happened to the velocity of the wave?<br />- it has increased.<br /><br />What is the amplitude of the wave now?<br />- 0.5 cm<br /><br />What is the wavelength of the wave now? <br />- 10 cm.<br /><br />What do you notice about the amplitude and wavelength of the wave as the velocity changes?<br />- amplitude and wavelength stay the same.<br /><br />Using this information, why do you think you see a lightning flash before hearing thunder?<br /><br />- because light travels much faster than sound.<br /><br />Click pause and find the minimum wavelength value.<br />- 5 cm<br /><br />What is the maximum wavelength value?<br />- 20 cm.<br /><br />How do they move?<br />- back and forth either side of their rest position.<br /><br />As a result of the passage of the wave, how have the layers either side of this one been moved from their rest position on one of the vertical lines?<br />- they have moved closer to this layer.<br /><br />As a result of the passage of the wave, how have the layers either side of this one been moved from their rest position on one of the vertical lines?<br />- they have moved further from this layer.<br /><br />From the graph, what is the approximate displacement of the green layers at the middle of a compression? <br />- zero<br /><br />From the graph, what is the approximate displacement of the red layers at the middle of a rarefaction?<br />- zero<br /><br />when the displacement graph is positive, which way have the particles been displaced from their rest positions?<br />- to the right.<br /><br />When the displacement graph is negative, which way have the particles been displaced from their rest positions?<br />- to the left.<br /><br />How does this affect the air pressure?<br />- it reduces it.<br /><br />What is the height in divisions of the crest of the wave on the screen?<br />- two divisions.<br /><br />What is the amplitude of the wave, in volts?<br />- 2 x 0.2 = 0.4 v<br /><br />What is the wavelength, in divisions, of the wave on the screen?<br />- four divisions.<br /><br />What time is represented by the wavelength value?<br />- 4 x 0.5 = 2 ms = 2/1000 in seconds<br /><br />Use this equation to calculate the frequency of the sound you see being displayed on the oscilloscope screen. <br />- 1000/2 = 500 hz.<br /><br />What is the periodic time and the frequency of this sound?<br />- about 12 ms and 80 hz.<br /><br /><br /><br />Combining Waves<br /><br />Vince Nguyen<br />Oscar Osorio<br />Lucas Liete<br />Abdou Williams<br />Kelvine Clarke<br /><br />How do their amplitudes and wavelengths compare?<br />- they are the same.<br /><br />Are the waves in phrase or out of phrase with each other?<br />- in phase.<br /><br />How do these two waves compare in amplitude and wavelength?<br />- They are the same.<br /><br />Are they in phase or out of phase with each other?<br />- theya re out of phase. The crests of one wave line up with the troughs of the other.<br /><br />Predict what will happen when two waves are added.<br />- they will cancel out.<br /><br />How do the two waves compare in amplitude?<br />- the amplitude of the blue wave is one third of the amplitude of the red wave.<br /><br />How do they compare in wavelength?<br />- the wavelength of the blue wave is one third of the wavelength of the red wave.<br /><br />How do they compare in frequency?<br />- the frequency of the blue wave is three times the frequency of the red wave.<br /><br />How far apart are the nodes or antinodes in terms of red or blue wavelengths?<br />- half a wavelength.<br /><br />When the red and blue waves are in phase are they added constructively or destructively?<br />- Constructively<br /><br />How many times does this resonance happen?<br />- four<br /><br />What are the frequencies?<br />- 15, 20, 45, and 60 units.<br /><br />How many antinodes are seen at each of these frequencies?<br />- one, two , three, and four antinodes, respectively.<br /><br />What is the ration of the frequencies that cause resonance?<br />- 1 3 5<br /><br />What is the ratio of these frequencies to the lowest frequency found in the closed tube?<br />- 1 2 4 <br /><br />So, from the length of your first resonant position value, what is the wavelength of the sound you have heard?<br />- 4 x 0.215 = 0.860 m<br /><br />From this equation and your wavelength value, what is the velocity of the sound in the air of the tube for this frequency of tuning fork?<br />- 384 x 0.860 = 330 m s ^-1<br /><br />How does this length compare with the first one?<br />- it is three times as long.<br /><br />Repeat the experiment and see how many resonant positions you can find. How many are there?<br />- three.<br /><br />From the length of the second one, where there are three node- antonode distances, what is the wavelength and the velocity of the sound?<br />- 0.688 m and 330 m s ^-1<br /><br />What happens to the wave?<br />- the wave spreads around the edges of the two obstacles.<br /><br />What difference do you see in the waves that diffract through the narrow gap?<br />- the waves are diffracted through the wider angle when they pass through a narrow gap.<br /><br />What do younotice about the area to the right of the barrier compared to the single – slit picture?<br />- these are two regions where the wave crests are missing.<br /><br />What would you expect to happen where two wave crests meet?<br />- a very high crest would be formed.<br /><br />Now look for places where a crest from one wave source meets a through from the other. What would you expect to happen in these places?<br />- the crest of one wave is cancelled out by the through of the other one, leaving the water calm.<br /><br />What would you expect to happen where two wave throughs meet?<br />- a very low through would formed.<br /><br />Now look for places where a crst from one wave source meets a through from the other. What would you expect to happen in these places?<br />- the crest of one wave is cancelled out by the trough of the other one, leaving the water calm.<br /><br />What happens at point marked with a red dot?<br />- a wave crest from source 1 has meet a crest from source 2 so they reinforce and make a high crest.<br /><br />What is the different between your two answers in terms of wavelengths?<br />- zero wavelengths.<br /><br />How do these differ from the other cases? <br />- they are both cancellations, where a crest from one source meets a through from the other.<br /><br />What do these straight lines represent?<br />- regions of cancellation where crests from one source meet throughs from the other.<br /><br />What does this do to x?<br />- increasing the source separation a decreases x.<br /><br />what does this do this x?<br />- increasing the wavelength increases x.<br /><br />what does this do to x?<br />- increasing the distance to the cross section sections increases x.<br />May 14, 2009 10:39 AMDavehttps://www.blogger.com/profile/14809267838191379368noreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-39152615112151753612009-05-15T09:40:00.000-04:002009-05-15T09:40:00.000-04:00Introducing Oscillations
• The cannonball is no...Introducing Oscillations<br /><br /><br /><br />• The cannonball is no longer experiencing an accelerating force. Acceleration is zero and velocity (neglecting air resistance) is constant<br /><br />• The force is independent of displacement<br />• The graph is a straight line that passes through the origin, showing that force is proportional to acceleration. The relationship F=ma applies to all motion of any constant mass, and it shows that the gradient of the graph is the mass m<br /><br />• When the force is zero, the displacement is also zero<br />• Force is proportional to displacement, but in the opposite direction<br />• The relationship is the same as before (but the maximum values of force and displacement are smaller).<br /><br />• When the acceleration is zero, the displacement is also zero<br />• The relative directions of force and acceleration are always in the same direction<br />• Force is proportional to acceleration<br />• When the displacement is zero, the acceleration is also zero at its maximum value<br />• The relative directions of acceleration and displacement are always in the opposite direction<br /><br />• The gradient of the graph has increased<br />• The acceleration is constant while the ball is in the barrel and velocity increases steadily while the ball is in the barrel<br /><br />• The force on the cannonball is constant and in the same direction as the displacement. A force producing oscillation varies with displacement and is in the opposite direction to it thus acting as a restoring force<br /><br />• Displacement and velocity are not always in the opposite directions<br />• When velocity has its maximum value, displacement is at zero<br />• The boxing glove is at the equilibrium position when it exhibits its fastest movement<br /><br />• When velocity is zero, the value of acceleration is at its maximum<br />• When displacement has its maximum values, the velocity is zero<br />• When the displacement is zero, the acceleration is also zero<br />• The values of 2(3.14)ft angle theta are the same<br />• Sin 2(3.14)ft may have values that ranges from negative 1 to positive 1<br />• X may have values that ranges from negative A to positive A<br /><br /><br />Wave Behavior<br /><br /><br /><br />• The reflected waves are moving downwards, across the original wave.<br />• To make the rippled tank produce waves of a smaller wavelength you must make the motor run faster so that the wooden beam vibrates more frequently.<br />• After reflection, the plane wave becomes a circular wave.<br />• The curved reflections waves get smaller and concentrate into a point.<br />• The reflection wave has a circular shape.<br />• The reflected waves seem to come from a point behind the reflector that is the same distance from it as the original source of the waves.<br />• The wave changes direction when it enters the shallow water.<br />• Wave speed is less in the shallow water.<br />• The wave is moving from left to right.<br />• The particles are moving up and down.<br />• The wave above the black centre line one cm.<br />• The distance between two wave tops is 15 cm.<br />• The amplitude of the wave is 5 cm.<br />• When you move the wavelength slider to the far left the wave crests get closer together.<br />• The wavelength is now 10 cm.<br />• The wavelength is now 20 cm.<br />• The amplitude is now 5 cm.<br />• The energy carried by the wave has not changed because the wave amplitude has not changed.<br />• The amplitude of the wave is 5cm and its wavelength is 15cm.<br />• The velocity of the wave has increased.<br />• No, the amplitude of the wave and wavelength did not change.<br />• The amplitude changes the height of the crest and the depths of the troughs, but not the wavelength or velocity.<br />• The wavelength changes the distance between neighboring crests and neighboring troughs, but not the amplitude or velocity.<br />• The other red particle is at the same height.<br />• The yellow particle is at its lowest point.<br />• The other red particle is at its lowest point.<br />• The yellow particle is at its highest point.<br />• The two red particles are one wavelength apart.<br />• The red and yellow particles are half a wavelength apart.<br />• The vertical ones get through both of the circular filters.<br />• The vertical waves cannot pass through the filter.<br />• The horizontal waves get through the first filter but not the second one.<br />• The horizontal waves can get through the second filter by rotating is at 90 degrees.<br />• The wave travelled 24m.<br />• The connection between the numbers in each row of the table is velocity = frequency x wavelength.<br />• The wave is moving from left to right.<br />• Their disturbance is different to the way they are distributed by a transverse wave because the particles are disturbed from left to right, rather than up and down.<br />• The maximum displacement of a particle from the middle of its motion is 0.5cm.<br />• The amplitude is about 0.25cm.<br />• The amount of energy was increased by increasing the amplitude.<br />• The distance between the centres of two neighboring compressions is approximately 10cm.<br />• The distance between the centres of two neighboring rarefactions is approximately 10cm.<br />• This compares to the distance between compressions because they are the same.<br />• The wavelength is about 13cm.<br />• The amplitude of the wave is 0.5cm.<br />• The wavelength of the wave is 10cm.<br />• The velocity of the wave has increased.<br />• The amplitude of the wave is 0.5cm.<br />• The wavelength of the wave is 10cm.<br />• As the velocity changes the amplitude and the wavelength stay the same.<br />• You see lightning flash before hearing thunder because light travels much faster than sound.<br />• The minimum wavelength value is 5cm.<br />• The maximum wavelength value is 20cm.<br />• The vertical layer of air particles move back and forth either side of their rest position.<br />• They moved closer to this layer.<br />• They have moved further from this layer.<br />• The approximate displacement of the green layers at the middle of a compression is zero.<br />• The approximate displacement of the red layers at the middle of a compression is zero.<br />• When the displacement graph is positive, the particles displace to the right.<br />• When the displacement graph is negative, the particles displace to the left.<br />• When the amplitude slider is set to a lower setting, the air pressure reduces.<br />• The height of the crest of the wave on the screen is two divisions.<br />• The amplitude of the wave in volts is 2 x 0.2 = 0.4V<br />• The wavelength of the wave on the screen is four divisions.<br />• The time represented by the wavelength value is 4 x 0.5 = 2/1000 in secondsAjaque, Jonathon, Damynicque, Martinanoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-46933852947003815332009-05-14T10:26:00.000-04:002009-05-14T10:26:00.000-04:00Combining Waves
How do their amplitudes and wavel...Combining Waves<br /><br />How do their amplitudes and wavelengths compare? same.<br /><br />Are the waves in phase or out of phase with each other? In phase.<br /><br />How do these two waves compare in amplitude and wavelength? the same.<br /><br />Are they in phase or out of phase with each other? They are out of phase. <br /><br />Predict what will happen when the two waves are added. They cancel out.<br />How do the two waves compare in amplitude? The amplitude of the blue wave is one third of the amplitude of the red wave.<br />How do they compare in wavelength? The wavelength of the blue wave is one third of the wavelength of the red wave.<br />How do they compare in frequency? The frequency of the blue wave is three times the frequency of the red wave.<br />How far apart are the nodes or antinodes in terms of red or blue wavelengths? Half of a wavelength.<br />When the red and blue waves are in phase are they added constructively or destructively? Constructively<br />When the red and blue waves are out of phase are they added constructively or destructively? Destructively<br />How many times does this resonance happen? 4<br />What are these frequencies? 15, 30, 45, and 60 units.<br />How many antinodes are seen at each of these frequencies? One, two, three, and four antinodes, respectively.<br />What is the ratio of the frequencies that cause resonance? 1:3:5<br />What is the ratio of these frequencies to the lowest frequency found in the closed tube? 1:2:4<br />So, from the length of your first resonant position, what is the wavelength of the sound you have heard? 4 × 0.215 = 0.860 m<br />From this equation and your wavelength value, what is the velocity of the sound in the air of the tube for this frequency of tuning fork? 384 × 0.860 = 330 m s–1<br />How does this length compare with the first one? Its 3 times as long.<br />Repeat the experiment and see how many resonant positions you can find. How many are there? 3<br />From the length of the second one, where there are three node-antinode distances, what is the wavelength and velocity of the sound? 0.688 m and 330 m s–1.<br />What happens to the wave? Spreads around two obstacles<br />What difference do you see in the waves that diffract through the narrow gap? The waves are diffracted through a wider angle when they pass through a narrow gap.<br />What do you notice about the area to the right of the barrier compared to the single-slit picture? The wave crests is missing in two regions.<br />What would you expect to happen where two wave crests meet? A high crest would be formed<br />What would you expect to happen where two wave troughs meet? A low trough would be formed<br />Now look for places where a crest from one wave source meets a trough from the other. What would you expect to happen in these places? The water is calm cause the crest of one wave is cancelled out by the trough of the other one<br />What happens at the point marked with a red dot? The wave crest from source 1 meets the wave crest from source 2 and makes a higher crest<br />What is the difference between your two answers in terms of wavelengths? Zero wavelengths.<br />How do these differ from the other cases? They are both cancellations, where a crest from one source meets a trough from the other.<br />What do these straight lines represent? Regions of cancellation where crests from one source meet troughs from the other.<br />What does this do to x? Increasing the source separation a decreases x.<br />What does this do to x? Increasing the wavelength increases x.<br />What does this do to x? Increasing the distance to the cross-section increases x.NikitaCecoyFreddyWagnernoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-10243308603441996882009-05-14T10:16:00.000-04:002009-05-14T10:16:00.000-04:00Slide 3
Which way are the reflected waves moving?...Slide 3<br /><br />Which way are the reflected waves moving?<br /><br />Downwards, across the original wave.<br /><br />How would you make the ripple tank produce waves of a smaller wavelength?<br /><br />Make the motor run faster so that the wooden beam vibrates more frequently.<br /><br /><br />Slide 4<br /><br />What happens to the plane wave after reflection?<br /><br />It becomes a circular wave.<br /><br />What happens to the curved reflected waves?<br /><br />They get smaller and concentrate into a point.<br /><br /><br />Slide 5<br /><br />What shape does the reflected wave have?<br /><br />A circular shape.<br /><br />Where do the reflected waves seem to come from?<br /><br />They seem to come from a point behind the reflector that is the same distance from it as the original source of the waves.<br /><br /><br />Slide 6<br /><br />What happens to the wave?<br /><br />The wave changes direction when it enters the shallow water.<br /><br />What happens to the speed of the waves?<br /><br />The wave speed is less in the shallow water.<br /><br /><br /><br />What happens to the wavelength?<br /><br />The wavelength is reduced in the shallow water.<br /><br /><br />Slide 8<br /><br />Which way is the wave moving?<br /><br />From left to right.<br /><br />Which way are the particles moving?<br /><br />Up and down.<br /><br />Assuming the background lines are 1 cm apart, how high is your wave above the black centre line?<br /><br />1 cm<br /><br />What is the amplitude now?<br /><br />7.5 cm<br /><br /><br />Slide 9<br /><br />Assuming the background lines are approximately 1 cm apart, what is the distance between two wave tops?<br /><br />15 cm<br /><br />What is the amplitude of the wave?<br /><br />5 cm<br /><br />Move the Wavelength slider to the far left. What happens?<br /><br />The wave crests get closer together.<br /><br />What is the wavelength now?<br /><br />10 cm<br /><br /><br /><br />What is the wavelength now?<br /><br />20 cm<br /><br />What is the amplitude now?<br /><br />5 cm<br /><br />Have you changed the energy carried by the wave? Explain your answer<br /><br />No, because the wave amplitude has not changed.<br /><br /><br /><br />Slide 10<br /><br />What is the amplitude of the wave and what is its wavelength?<br /><br />Its amplitude is 5 cm and its wavelength is 15 cm.<br /><br />What has happened to the velocity of the wave?<br /><br />It has increased.<br /><br />Measure the amplitude of the wave and wavelength again. Have they changed?<br /><br />No<br /><br /><br />Slide 12<br /><br />Move the Amplitude slider back and forth to see its effect on all parts of the wave. What happens?<br /><br />The amplitude changes the height of the crests and the depth of the troughs, but not the wavelength or velocity.<br /><br />Move the Wavelength slider back and forth to see its effect on all parts of the wave. What happens?<br /><br />The wavelength changes the distance between neighboring crests and neighboring troughs, but not the amplitude or velocity.<br /><br /><br />Slide 13<br /><br />Where is the other red particle at this moment?<br />At the same height.<br />Where is the yellow particle at this moment?<br />At its lowest point.<br />Where is the other red particle at this moment?<br />At its lowest point.<br />Where is the yellow particle at this moment?<br />At its highest point.<br />How far apart are the two red particles?<br />One wavelength.<br />How far apart are the red and yellow particles?<br />Half a wavelength.<br /><br />Slide 14<br />Which ones get through both of the circular filters? <br />The vertical ones.<br />What happens to the vertical waves now?<br />They cannot pass through the filter.<br />What happens to the horizontal waves now? <br />They get through the first filter but not the second one.<br />How could you allow the horizontal waves to get through the second filter?<br />By rotating it 90°.<br /><br />Slide 15<br />Look at the length of the wave on the scale. How far did the wave travel?<br />24 m<br />What is the connection between the numbers in each row of the table?<br />Velocity = frequency × wavelength<br /><br />Slide 17<br />Which way is the wave moving? <br />From left to right.<br />Look at the particles in the material. How is their disturbance different to the way they are disturbed by a transverse wave? <br />Here the particles are disturbed from left to right, rather than up and down.<br />Assuming the background lines are 1 cm apart, what is the maximum displacement of a particle from the middle of its motion? <br />0.5 cm<br />What is the amplitude now? <br />About 0.25 cm.<br />How did you do this?<br />By increasing the amplitude.<br /><br />Slide 18<br />Assuming the vertical background lines are 1 cm apart, what is the distance between the centres of two neighboring compressions? (Particles at the centre of each compression should be on, or close to a vertical line.)<br />Approximately 10 cm.<br />What is the distance between the centres of two neighboring rarefactions? (Again, particles at the centre of each rarefaction should be on, or close to a vertical line.)<br />Approximately 10 cm.<br />How does this compare to the distance between compressions?<br />It is the same.<br />Pause the wave. What is the wavelength now? <br />About 13 cm.<br /><br />Slide 19<br />What is the amplitude of the wave? <br />0.5 cm<br />What is the wavelength of the wave? <br />10 cm<br />What has happened to the velocity of the wave? <br />It has increased.<br />What is the amplitude of the wave now? <br />0.5 cm<br />What is the wavelength of the wave now? <br />10 cm<br />What do you notice about the amplitude and wavelength of the wave as the velocity changes?<br />Amplitude and wavelength stay the same.<br />Why do you think you see a lightning flash before hearing thunder?<br />Because light travels much faster than sound.<br /><br />Slide 20<br />Find the minimum wavelength value.<br />5 cm<br />What is the maximum wavelength value?<br />20 cm<br /><br />Slide 22<br />How do they move?<br />Back and forth either side of their rest position.<br />As a result of the passage of the wave, how have the layers either side of this one been moved from their rest position on one of the vertical lines?<br />They have moved closer to this layer.<br />As a result of the passage of the wave, how have the layers either side of this one been moved from their rest position on one of the vertical lines?<br />They have moved further from this layer.<br /><br />Slide 23<br />From the graph, what is the approximate displacement of the green layers at the middle of a compression? <br />Zero<br />From the graph, what is the approximate displacement of the red layers at the middle of a rarefaction? <br />Zero<br /><br />When the displacement graph is positive, which way have the particles been displaced from their rest positions? <br />To the right.<br />When the displacement graph is negative, which way have the particles been displaced from their rest positions?<br />To the left.<br /><br />Slide 24<br />How does this affect the air pressure?<br />It reduces it.<br /><br />Slide 25<br />What is the height in divisions of the crest of the wave on the screen? <br />Two divisions.<br />What is the amplitude of the wave, in volts? <br />2 × 0.2 = 0.4 V<br />What is the wavelength, in divisions, of the wave on the screen? <br />Four Divisions<br />What time is represented by the wavelength value? <br />4 X 0.5 = 2ms = 2/1000 in seconds<br />Use this equation to calculate the frequency of the sound you see being displayed on the oscilloscope screen.<br />1000/2 = 500Hz<br />What is the periodic time and the frequency of this sound? <br />About 12 ms and 80 Hz<br /><br />Lesson3<br />How do their amplitudes and wavelengths compare?<br /> They are the same.<br />Are the waves in phase or out of phase with each other?<br /> In phase.<br />How do these two waves compare in amplitude and wavelength?<br /> They are the same.<br />Predict what will happen when the two waves are added<br /> They will cancel out.<br />How do the two waves compare in amplitude?<br /> The amplitude of the blue wave is one third of the amplitude of the red wave.<br />How do they compare in wavelength?<br /> The wavelength of the blue wave is one third of the wavelength of the red wave.<br />How do they compare in frequency?<br /> The frequency of the blue wave is three times the frequency of the red wave.<br />How far apart are the nodes or antinodes in terms of red or blue wavelengths?<br /> Half a wavelength.<br />When the red and blue waves are in phase are they added constructively or destructively?<br /> Constructively<br />When the red and blue waves are out of phase are they added constructively or destructively?<br /> Destructively<br />How many times does this resonance happen?<br /> Four<br />What are these frequencies?<br /> 15, 30, 45, and 60 units.<br />How many antinodes are seen at each of these frequencies?<br /> One, two, three, and four antinodes, respectively.<br />What is the ratio of the frequencies that cause resonance?<br /> 1:3:5<br />What is the ratio of these frequencies to the lowest frequency found in the closed tube?<br /> 1:2:4<br />So, from the length of your first resonant position, what is the wavelength of the sound you have heard?<br /> 4 × 0.215 = 0.860 m<br />From this equation and your wavelength value, what is the velocity of the sound in the air of the tube for this frequency of tuning fork?<br /> 384 × 0.860 = 330 m s–1<br />How does this length compare with the first one?<br /> It is three times as long.<br />How many resonant positions you can find. How many are there?<br /> Three<br />What is the wavelength and velocity of the sound?<br /> 0.688 m and 330 m s–1.<br />What happens to the wave?<br /> The wave spreads around the edges of the two obstacles.<br />What difference do you see in the waves that diffract through the narrow gap?<br /> The waves are diffracted through a wider angle when they pass through a narrow gap.<br />What do you notice about the area to the right of the barrier compared to the single-slit picture?<br /> There are two regions where the wave crests are missing.<br />What would you expect to happen where two wave crests meet?<br /> A very high crest would be formed.<br />What would you expect to happen where two wave troughs meet?<br /> A very low trough would be formed.<br />What would you expect to happen in these places?<br /> The crest of one wave is cancelled out by the trough of the other one, leaving the water calm.<br />What happens at the point marked with a red dot?<br /> A wave crest from source 1 has met a crest from source 2 so they reinforce and make a high crest.<br />What is the difference between your two answers in terms of wavelengths?<br /> Zero wavelengths.<br />How do these differ from the other cases?<br /> They are both cancellations, where a crest from one source meets a trough from the other.<br />What do these straight lines represent?<br /> Regions of cancellation where crests from one source meet troughs from the other.<br />What does this do to x?<br /> Increasing the source separation a decreases x.<br />What does this do to x?<br /> Increasing the wavelength increases x.<br /><br />What does this do to x?<br /> Increasing the distance to the cross-section increases x.Natalie Thepkaysone, Marlon Tevesnoreply@blogger.comtag:blogger.com,1999:blog-4107303980700785197.post-55903432872141204202009-05-14T10:09:00.000-04:002009-05-14T10:09:00.000-04:00Wave Behavior
1. Which way are the reflected wave...Wave Behavior<br /><br />1. Which way are the reflected waves moving? They are moving down and across the original wave.<br />2. How would you make the ripple tank produce waves of a smaller wavelength? You would have to make the motor move faster to make the waves move more frequently<br />3. What happens to plane waves after reflection? It turns into a circular wave.<br />4. What happens to the curved reflected waves? They start to become smaller and focus on to one point.<br />5. What shape does the reflected wave have? circular<br />6. Where do the reflected waves seem to come from? They come from a point behind the reflector which is the same distance from the original source of the waves.<br />7. What happens to the wave? when entering the shallow water the waves start to change direction..<br />8. What happens to the speed of the waves? The wave speed is slower.<br />9. What happens to the wavelength? The wavelength is reduced.<br />10. Which way is the wave moving? left to right.<br />11. Which way are the particles moving? Up and down.<br />12. How high is your wave above the center line? 1 cm<br />13. What is the amplitude now? 7.5 cm<br />14. What is the distance between two wave tops? 15 cm<br />15. What is the amplitude of the wave? 5 cm<br />16. What happens? The wave crests get closer together.<br />17. What is the wavelength now? 10 cm<br />18. What is the wavelength now? 20 cm<br />19. What is the amplitude now? 5 cm<br />20. Have you changed the energy carried by the wave? the wave amplitude did not change.<br />21. What is the amplitude of the wave and its wavelength? Its amplitude is 5 cm and its wavelength is 15 cm.<br />22. What happened to the velocity of the wave? It increased<br />23. Have they changed? No<br />24. What happens? the wavelength and velocity remain the same but the height of the crests and the depth of the troughs were changed<br />25. What happens? the distance between neighboring crests and neighboring troughs were changed but not the amplitude or velocity.<br />26. Where is the other red particle at that moment? Same height<br />27. Where is the yellow particle at this moment? At the lowest point<br />28. Where is the other red particle at this moment? Lowest point<br />29. Where is the yellow particle at this moment? Highest point<br />30. How far apart are the two red particles? One wavelength<br />31. How far apart are the red and yellow particles? Half a wavelength<br />32. Which ones get through both the circular filter? The vertical ones<br />33. what happens to the vertical waves now? They cannot pass the filter<br />34. what happens to the horizontal waves now? They get through the first but not the second<br />35. how could you allow the horizontal waves to get throught the second filiter? rotating it 90 degrees<br />36. How far did the wave travel? 24 m<br />37. What is the connection between the numbers in each row of the table? v= fw<br />38. Which way is the wave moving? left to right.<br />39. Look at the particles in the material. How is their disturbance different to the way they are disturbed by a transverse wave? the particles are moved from left to right, rather than up and down.<br />40. Assuming the background lines are 1 cm apart, what is the maximum displacement of a particle from the middle of its motion? 0.5 cm<br />41. What is the amplitude now? About 0.25 cm.<br />42. How did you do this? increasing the amplitude.<br />43. Assuming the vertical background lines are 1 cm apart, what is the distance between the centres of two neighbouring compressions? Approximately 10 cm. <br />44. Approximately 10 cm. Approximately 10 cm. <br />45. How does this compare to the distance between compressions? the same. <br />46. What is the wavelength now? About 13 cm. <br />47. What is the amplitude of the wave? 0.5 cm <br />48. What is the wavelength of the wave? 10 cm <br />49. What has happened to the velocity of the wave? Increased. <br />50. What is the amplitude of the wave now? 0.5 cm <br />51. What is the wavelength of the wave now? 10 cm <br />52. What do you notice about the amplitude and wavelength of the wave as the velocity changes? stays the same. <br />53. Using this information, why do you think you see a lightning flash before hearing thunder? Light travels faster than sound. <br />54. find the minimum wavelength value. 5 cm <br />55. What is the maximum wavelength value? 20 cm <br />56. How do they move? Back and forth either side of their rest position. <br />57. As a result of the passage of the wave, how have the layers either side of this one been moved from their rest position on one of the vertical lines? moved closer to this layer. <br />58. As a result of the passage of the wave, how have the layers either side of this one been moved from their rest position on one of the vertical lines? moved farther from this layer. <br />59. From the graph, what is the approximate displacement of the green layers at the middle of a compression? Zero <br />60. From the graph, what is the approximate displacement of the red layers at the middle of a rarefaction? Zero<br />61. When the displacement graph is positive, which way have the particles been displaced from their rest positions? To the right.<br />62. When the displacement graph is negative, which way have the particles been displaced from their rest positions? To the left.<br />63. How does this affect the air pressure? reduces it.<br />64. What is the height in divisions of the crest of the wave on the screen? Two divisions. <br />65. What is the amplitude of the wave, in volts? 2 × 0.2 = 0.4 V<br />66. What is the wavelength, in divisions, of the wave on the screen? Four divisions.<br />67. What time is represented by the wavelength value? 4 × 0.5 = 2 ms = 2/1000 in seconds<br />68. Use this equation to calculate the frequency of the sound you see being displayed on the oscilloscope screen. 1000/2 = 500 Hz<br />69. What is the periodic time and the frequency of this sound? About 12 ms and 80 Hz.NikitaCecoyFreddyWagnernoreply@blogger.com