Calculating Mass of a Block in SHM with Given Parameters

[SpyIsCake]

Homework Statement

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

Homework Equations

T = 2π√(m/k)
T = d/v

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.

 

[Chestermiller]

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

 

[SpyIsCake]

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

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

 

[Chestermiller]

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

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

Chet

 

[mfb]

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

 

[SpyIsCake]

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

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

 

[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

Whoops, I actually got a mass of 0.37kg.

Still not the answer, though.

 

[mfb]

The energy stored in a spring at 0.15m[/color] is:

1/2 (75N/m)(0.12[/color]^2) = 0.54 J of energy.

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.

 

[SpyIsCake]

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.

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.

 

[mfb]

I don't have speed at 0.15m.

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

 

[SpyIsCake]

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

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.

 

[mfb]

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

 

[SpyIsCake]

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

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.

 

[Chestermiller]

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

 

[mfb]

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.

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