# Vertical SHM

1. May 22, 2005

### WY

Hi
I'm attempting this question and I'm wondering if this is the correct way of going about it or if i'm completely off track:
For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italian ham. The slices of ham are weighed on a plate of mass 0.400 kg placed atop a vertical spring of negligible mass and force constant of 200 N/m. The slices of ham are dropped on the plate all at the same time from a height of 0.250 m. They make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (SHM). You may assume that the collision time is extremely small.

What is the amplitude of oscillation A of the scale after the slices of ham land on the plate?
What is the period of oscillation T of the scale?

I started using conservation of GPE and equating it to the energy of the spring
mgh = 1/2*k*A^2
and made A the subject to find the amplitude - would this give me the right answer?

And then for the second part I used (2*pi)/sqrt(k/m)=T and m = the mass of ham - should i include the mass of the plate?

2. May 22, 2005

### whozum

No, the 'X' in hooke's law represents the total displacement, which is actually a multiple of the amplitude.

At a glance the second part looks fine. You need to include the mass of the scale since it is also part of the SHM.

3. May 22, 2005

### WY

a multiple of the amplitude - so would i have to divide it by 2 to find the amplitude?

4. May 22, 2005

### Staff: Mentor

No. Use conservation of energy to find out how far the scale depresses from its initial position after the downward motion stops. Don't forget to include how far the scale depresses in calculating the GPE. Once you've found the lowest point, compare that to the equilibrium position of the system to figure out the amplitude. (You'll have to find the new equilibrium position.)

5. May 22, 2005

### ussrasu

I still dont understand how to get it - can you explain it in detail?

6. May 22, 2005

### Staff: Mentor

I thought I just did?

(1) Call the initial vertical position of the plate 0 and its lowest point (after the ham hits it) x. Now write an expression for conservation of energy to find x.

(2) Before the ham hits the plate, the equilibrium of the system is at position 0. How far below zero is the new equilibrium position of the "plate + ham" system?

(3) Combine the answers from 1 and 2 to figure out the amplitude of the motion about the new equilibrium position.

7. May 22, 2005

### ussrasu

Sorry can you show how to find x - the lowest position (after the ham hits it)?

8. May 22, 2005

### Staff: Mentor

I describe how to do that as step 1 above. Try it! It's just conservation of energy. The ham starts with gravitational PE which gets transformed into spring PE. Note that the ham doesn't just lower a distance h = 0.250 m. (Measure the change in height of the ham from its initial position above the plate to the lowest point of the motion.)

9. May 22, 2005

### ussrasu

But then if mgh = 1/2kA^2
that means for the ham:
0.3*9.8*0.25 = 1/2*200*A^2?
I dont understand? Ive done that before but it doesnt work?
But what you are saying is to work out how far the ham goes past the 0.25m? I still dont understand how to get that?

10. May 22, 2005

### Staff: Mentor

That is incorrect, as I stated in my first post in this thread. It is incorrect for two reasons:

(1) The potential energy of the ham is not mgh, where h is the height above the plate. Measure the height of the ham from the lowest position. (The lowest position is what I call "x" below the original position. So... if the ham starts at a height "h" above the plate, and ends up "x" below the plate... what's the total change in height?)

(2) The displacement from the original position is not the amplitude! The spring potential energy is given by 1/2 k x^2, where x = how far the plate lowers from its original position.

Given these tips, see if you can write the correct statement of conservation of energy.

11. May 23, 2005

### ussrasu

So then U = mg(h+x)
therefore: 0.3*9.8*(0.25+x) = 1/2*200*x^2
So then do you solve for x, using the quadratic formula?
I get x = 0.455 or x = -0.16
Which one would then be the displacement?
I got these answers from: 100x^2 - 29.4x -7.35
Am i doing the right thing now?

12. May 23, 2005

### ussrasu

Im not sure if im doing the right thing here? Can anyone show me how this question is solved?

13. May 23, 2005

### ~angel~

Have you used the hints? Please don't tell me you're one of those people who refuse to use the hints =P

14. May 23, 2005

### OlderDan

Looks like you now have the potential energy of the ham relative to PE = zero at the stopping point. What about the plate?

15. May 23, 2005

### Staff: Mentor

Exactly!
Almost. You made an error when you plugged in the numbers. I think the quadratic that you need to solve should be: 100x^2 - 2.94x - 0.735 = 0

You'll get two answers, but only one is relevant to this problem (the positive one).

16. May 23, 2005

### Staff: Mentor

Representing the energy of the "plate + spring" system as $1/2 k x^2$ where x is measured from the equilibrium position (the original position before the ham drops) automatically includes the change in PE of the plate.

17. May 23, 2005

### OlderDan

Sorry. I lost track of what you meant by x, but even if it was the amplitude I would have been off by a term for failing to include the change in equilibrium position as part of the height of the ham relative to final equilibrium.

I've been having some problem resolving your statement, which I know to be true for a mass hanging on a spring. I keep getting inconsistent results depending on how I write the displacements. At first I thought it had to be an algebra mistake, but now I'm thinking there is a fundamental problem with the approach.

The problem clearly states that the collision is inelastic and that the collision time is very small (can be neglected). To me that suggests that the collision needs to be treated as an instantaneous conservation of momentum problem to find an initial velocity of the combined mass/spring system at a position above final equilibrium. Exteral forces can be neglected during the collison. After the collision, energy will be conserved, but not for the whole process. Does that make sense??

18. May 23, 2005

### Staff: Mentor

D'oh! I misread the problem as saying "totally elastic collision".

Yep.

Good catch, OlderDan. Thanks!

19. May 23, 2005

### Staff: Mentor

Allow me to correct the misinformation from my earlier post (post #6 in this thread).
As OlderDan points out, first calculate the speed of the "ham + plate" after the collision using conservation of momentum. Then you can apply conservation of energy. There are several ways to track the energy. The easiest may be to treat the "plate + spring" as a system initially at its equilibrium point.

Thus, the initial energy of the system (after the collision) is the sum of: (a) KE of the ham, (b) gravitational PE of the ham, (c) KE of the "plate + spring" system (just the KE of the plate), and (d) spring PE of the "plate + spring" system (which equals zero, since it's at equilibrium).

The final energy of the system (as it reaches the lowest point) is the sum of those same terms. Of course, the KE terms are zero, since the system is momentarily at rest at the lowest point.

Note that by measuring the spring PE of the "plate + spring" system from its equilibrium point (which is its initial position) as $1/2 k x^2$, the gravitational PE of the plate is already accounted for.

Still valid.

Still valid.

20. May 23, 2005

### ussrasu

So is this positive x value now the displacement after the ham hits the plate? Is it also the amplitude or is this just the new equilibrium position from where the plate and ham are at rest?