# Simple Harmonics Question

1. Dec 5, 2007

### kevinr

[SOLVED] Simple Harmonics Question

1. The problem statement, all variables and given/known data

A 1.40 kg ball and a 2.20 kg ball are glued together with the lighter one below the heavier one. The upper ball is attached to a vertical ideal spring of force constant 170 N/m, and the system is vibrating vertically with amplitude 19.0 cm. The glue connecting the balls is old and weak, and it suddenly comes loose when the balls are at the lowest position in their motion.

m1 = 1.4 kg
m2 = 2.2 kg
k = 170 (not sure)

2. Relevant equations

(1/2)kx^2
(1/2)mv^2
(m1+ m2)v(both) = mv.

3. The attempt at a solution

Ok so i got the frequency of this problem to be 1.4 and thats correct. But im lost in finding the amplitude.

Here's what i have done but its wrong. (I think since mass is less - amplitude should be less)

i want to use conservation of momentum (m(both)v(both) = mv) to find v of m2 so i first need v of both.

For that i need v(both) so i used conservation of energy (1/2kx^2 = 1/2mv^2) -> which gives me v(both)

With that i use to find v of m2 using m2v2 = m(both)v(both). Now with the v of m2, i use energy conservation again (1/2kx^2 = 1/2mv^2) to find x for just m2. This should be the amplitude but the answer seems to be wrong) - i get 17 cm.

Is my approach ok? Any help would be great.

Thanks!

2. Dec 5, 2007

### Staff: Mentor

What is it that you are asked to find?

3. Dec 5, 2007

### kevinr

Amplitude after the small ball has fallen off. (of the big mass only)

Oops forgot that

4. Dec 5, 2007

### Staff: Mentor

OK. Hint: Where's the new equilibrium point?

5. Dec 5, 2007

### kevinr

Ok so that means that the 17 cm i got is compared to the old equilibrium?

If so, i have no idea how to get the new equilibrium point. Wouldnt the new equilibrium point be the new altitude you get? (since altitude is measured from equilibrium).

Thanks for help!

6. Dec 5, 2007

### Staff: Mentor

I have no idea what you were doing when you calculated that value. You don't need conservation of momentum or energy to solve this.

You are given that the amplitude (before the piece falls off) was 19 cm. That means the system oscillated between 19 cm above the equilibrium point to 19 cm below the equilibrium point. (The equilibrium point is where the net force on the mass is zero.)

When the piece falls off, the remaining mass is 19 cm below the old equilibrium point. You need to find out how far it is below the new equilibrium point. So figure out how much the equilibrium point shifts.

7. Dec 5, 2007

### kevinr

removed cause i figured out after the hint!

Last edited: Dec 5, 2007
8. Dec 5, 2007

### kevinr

Thanks!

I managed to figure it out after you Hint =D.