# Amplitude of a spring

1. May 18, 2015

### goonking

1. The problem statement, all variables and given/known data
Lets say I have a spring, resting at a length x at t = 0 seconds, with a spring constant k, and at the end is attached with a block with mass m. I give the block a velocity v in the -x direction.

Now I'm asked to find the Amplitude.

2. Relevant equations

3. The attempt at a solution

Here, I'm thinking, KEmax = PEmax = 1/2 m v2 = 1/2 k A2

since the highest velocity is the initial 'push' i give it, is it safe to say I have every variable to solve for A(amplitude)?

2. May 18, 2015

### Staff: Mentor

Looks good to me so far...

3. May 18, 2015

### goonking

so when is 1/2 k A2 = 1/2 mv2 + 1/2 k x2?

is that when you don't know the max velocity? is that when you know a given velocity and distance at a given time?

4. May 18, 2015

### Staff: Mentor

This looks correct for when v = the velocity at zero displacement.

This is confusing. Why are you adding the two terms on the RHS?

5. May 18, 2015

### goonking

I think to add up the total energy at a given time so KEmax =PEmax = 1/2 k x2 at a given time + 1/2 mv2 at a given time

I could be wrong though

6. May 18, 2015

### Staff: Mentor

That does look correct, but you should add subscripts to the velocity v to make it clear what you mean in each case. KE max is at zero displacement, and PE max is at maximum displacement. And KE + PE at any given point is equal to KE max or PE max.

7. May 18, 2015

### goonking

KEmax =PEmax = 1/2 k x^2 at a given time + 1/2 mv^2 at a given time = 1/2 k A 2

then I can solve for Amplitude, if I'm given a time where displacement isn't 0

correct?

8. May 18, 2015

### Staff: Mentor

Not quite.

KEmax = PEmax = 1/2 k A^2 = 1/2 m v(x=0)^2. At any given time KEmax = PEmax = 1/2 k x^2 + 1/2 m v^2

Where A = max amplitude.

9. May 18, 2015

### goonking

isn't that what i typed out?

10. May 18, 2015

### Staff: Mentor

Not the way I read it. I bolded the things I changed and added.