# Homework Help: Oscillations homework

1. Jun 23, 2012

### amiras

1. The problem statement, all variables and given/known data
A M=1.5kg horizontal, uniform tray is attached to a vertical ideal spring of force constant k=185 N/m and m=275g metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down 15cm below its equilibrium position and released from rest.

2. Relevant equations
How high above the release position will the tray be when the ball leave the tray?

3. The attempt at a solution
Lets call distance from the release point to equilibrium point A. Imagine the tray pushing the ball from release point to equilibrium point, when this distance is reached the acceleration of the system instantaneously becomes zero and the spring is now about to pull the tray downwards. At this moment the ball is no longer pushes the tray down, however because the ball does not push the tray down the equilibrium position is no longer the same as before but is a little above.
So the tray is going to keep up same speed with the ball as long it reaches its new equilibrium position.

If my idea is right, then the extra distance kx = mg, and x = mg/k, and final answer should be A + x

Is that right?

2. Jun 23, 2012

### azizlwl

Re: Oscillations

I take it on conservation of energy.
The velocity increases from release.
The velocity reached when it is equal to initial PE and leave the tray.

3. Jun 23, 2012

### ehild

Re: Oscillations

You are on the right track, but think over what the equilibrium position is. The system is in equilibrium when the net force on tray and ball is zero. How much is the spring shorter than its original length? Does it depend on the mass of the ball only? And what is "A"?

The ball and tray interact with the normal force N. The normal force can only push the ball upward. The ball leaves the tray when the normal force becomes zero, that is the gravitational acceleration is the same as the acceleration of the tray. As gravity acts to the tray too, the elastic force on the tray is zero at that instant: the length of the spring is equal to the uncompressed length. From that instant, the tray has greater downward acceleration then the ball, as the elastic force points downward when the length becomes longer than the uncompressed length.

ehild

4. Jun 23, 2012

### amiras

Re: Oscillations

I dont really see how could you do it with energy conservation

5. Jun 23, 2012

### azizlwl

Re: Oscillations

Initial U equal to final U.

Last edited: Jun 23, 2012
6. Jun 23, 2012

### amiras

Re: Oscillations

I understand the point ehild was making, and now solved the problem using that method. Now lets get it with the energy conservation.
I see no direct way of computing the distance immediately using energy conservation, except that first finding the velocity of the ball at that instant and then then using another equations to solve for the displacement.

So since the spring is initially compressed distance x = (m+M)g/k, and later on compressed more distance of A = 0.15 m, the string is compressed distance A + x and,

Initial elastic potential energy 1/2 k(A+x)^2
Initial gravitational potential energ. 0
Initial Kinetic energy 0

Final elastic energy 0
Final Gravitational mg(A+x)
Final Kinetic 1/2 mv^2

Gives the correct value of velocity.

Thanks everybody for help.