# Homework Help: Find mass of block. SHM

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1. Oct 18, 2014

### SpyIsCake

1. The problem statement, all variables and given/known data
A block attached to a horizontal spring of force constant 75 N/m undergoes SHM with an amplitude of 0.15 m. If the speed of the mass is 1.7 m/s when the displacement is 0.12 m from the equilibrium position, what is the mass of the block?
k = 75 N
v = 1.7m/s @ 0.12m displacement

2. Relevant equations
T = 2π√(m/k)
T = d/v
3. The attempt at a solution
I am having a really hard time figuring this out.
This question would be much easier if I could find the velocity when it is at its highest amplitude, because then I would have to multiply the time by two to get the full cycle.

But it does not give me that. I have no clue how to find the period in order to complete my equation.

2. Oct 18, 2014

### Staff: Mentor

What is the equation for the displacement from the equilibrium position as a function of time in SHM?

3. Oct 18, 2014

### SpyIsCake

I don't know. I wrote down the equation for SHM that gives period.

4. Oct 19, 2014

### Staff: Mentor

That equation is not adequate to solve this problem. Please go back and review your notes on simple harmonic motion.

Chet

5. Oct 19, 2014

### Staff: Mentor

What is the energy stored in the spring at a displacement of 0.15m? What is the value at 0.12m?

6. Oct 19, 2014

### SpyIsCake

The energy stored in a spring at 0.15m is:

1/2 (75N/m)(0.12^2) = 0.54 J of energy. Now, I can substitute the equation and solve for m like this:

1/2(75N/m)(0.12^2) = 1/2(M)(1.7^2)

I get a mass of 0.6 kg.

Answer is 0.21 kg

7. Oct 19, 2014

### SpyIsCake

Whoops, I actually got a mass of 0.37kg.

Still not the answer, though.

8. Oct 19, 2014

### Staff: Mentor

Huh?
You should get two different energy values for both displacements. Just this difference is available as kinetic energy, not the full energy stored in the spring.

9. Oct 20, 2014

### SpyIsCake

That was a typo.

I meant to calculate the energy value for when the spring is at 0.12m. The reason is because I have the speed on that value. I don't have speed at 0.15m.

10. Oct 20, 2014

### Staff: Mentor

It is the maximal distance the object can get away from the origin. What do you expect as velocity there?

11. Oct 20, 2014

### SpyIsCake

That's what I'm confused about. If I have that velocity, I should be able to get the mass.

But I don't know that velocity, nor do I know how to calculate it.

12. Oct 21, 2014

### Staff: Mentor

Does it move to the left or to the right at that point?
Trick question...

13. Oct 21, 2014

### SpyIsCake

I'll just go ahead and assume that this SHM is a mass on a spring hanging from the ceiling. On the highest amplitude, it's not really moving too much.

14. Oct 21, 2014

### Staff: Mentor

The equation I was thinking of in post number #2 was $y=0.15sin(\sqrt{\frac{k}{m}}t)$. From this you can calculate the velocity at time t.

Chet

15. Oct 21, 2014

### Staff: Mentor

It is a horizontal spring and horizontal motion. Anyway, the mass is just turning around, so it has a velocity of zero there.