# USAPhO 2009 F=MA exam #13, (spring SHM)

1. Jan 25, 2014

### Agrasin

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

A mass is attached to the wall by a spring of constant k. When the spring is at its natural length,
the mass is given a certain initial velocity, resulting in oscillations of amplitude A. If the spring
is replaced by a spring of constant 2k, and the mass is given the same initial velocity, what is the
amplitude of the resulting oscillation?

2. Relevant equations
Uh... This is the problem here. I might just be forgetful, but I cannot recall any equation relating amplitude and f, T, k, or v.

T = 2∏√(m/k) is the most relevant formula I could recall.

3. The attempt at a solution

Intuition led me to choose the correct answer: (1/√2)A

I know if k is doubled, the amplitude will decrease. I had a feeling there is a relationship k ∝ A^2 given a constant v_initial and mass. Can anyone shed some light on this?

2. Jan 25, 2014

### haruspex

Think about KE and PE. What did doubling the spring constant do to the total energy of the system?

3. Jan 26, 2014

### Agrasin

PE = (kx2)/2

Doubling k doubles the PE at any point x.

Thank you. But now I'm trying to think of an equation relating PE and Amplitude and I can't think of anything.

4. Jan 26, 2014

### voko

What is amplitude?

5. Jan 26, 2014

### Agrasin

Voko, I don't know what you're asking. If you want the definition of amplitude, it is the distance (in this case) from equilibrium to the maximum compression/ extension of the spring.

I'm still thinking of an equation relating PE with amplitude.

6. Jan 26, 2014

### voko

So amplitude has something to do with distance. So does PE. Can you connect the two?

7. Jan 26, 2014

### Agrasin

Oh wow! I must have been suffering from a massive brain fart.

PE = (kx2)/2

Given the same kinetic energy (v stays constant), and a doubled spring constant, PE1 = PE2 = (2k(x/√2)2)/2

x2 = (1/√2)A
Thanks.