Vertical Spring Oscillation Frequency

[jzwiep]

Homework Statement

A spring is standing upright on a table with its bottom end fastened to the table. A block is dropped from a height 2.8 cm above the top of the spring. The block sticks to the top end of the spring and then oscillates with an amplitude of 12 cm.

What is the oscillation frequency? (Hz)

Homework Equations

Don't know.

The Attempt at a Solution

The only thing I can think of is it looks like a conservation of energy question, but I don't know where to start.

Can anyone give me a nudge in the right direction?

 

[ideasrule]

What's the equation for the frequency of a mass-spring system? Hint: it includes m and k. You need to figure out k based on the amplitude.

 

[jzwiep]

Okay, so I found the formula relating k to f:

f=(1/2pi)sqrt(k/m)

I haven't been able to find anything that would relate k to A without any additional variables (like the vmax formula in terms of k, m, and A).

Whatever it is has to cancel out the mass in the f formula, but is there more info in the question that I'm just not getting? Would the fact that it is released 2.8cm above the top of the spring give me another piece of information?

Thanks.

 

[jzwiep]

Alright, I tried using conservation of energy.

U=mg(A+dL+h)

Where A is the amplitude, dL is the spring compression to equilibrium, and h is the height the height the mass was dropped above the spring.

Along with:

Fg=Fs
mg=k*dL

and

f=(1/2pi)sqrt(k/m)Am I at least on the right track? I can't seem to find an answer, so some help would be much appreciated.

 

[Doc Al]

Alright, I tried using conservation of energy.

U=mg(A+dL+h)

You're on the right track. Where's your conservation of energy equation?

 

[jzwiep]

Etotal=Us=Up
1/2kA^2=mg(A+dL+h)

I rearrange with equilibrium forces equation to get:

6k^2-145mk-96m^2=0

Then using the quadratic equation to yield:

(145m +/- 220.62m)/144

So k=2.539m

Then I substituted in for k in

f=(1/2pi)sqrt(k/m)

f=(1/2pi)sqrt(2.539)
=3.9883Hz
=4.0Hz

Which wasn't the right answer.

 

[Doc Al]

Etotal=Us=Up
1/2kA^2=mg(A+dL+h)

Careful here. The spring is compressed from its original uncompressed position, not just from the new equilibrium position.

 

[jzwiep]

Etotal=Us=Up
1/2k(A+dL)^2=mg(A+dL+h)

I rearrange with equilibrium forces equation to get:

72k^2-145mk-48m^2=0

Then using the quadratic equation to yield:

(145m +/- 186.68m)/144

So k=2.303m

Then I substituted in for k in

f=(1/2pi)sqrt(k/m)

f=(1/2pi)sqrt(2.303)
=2.38Hz

Is this now the correct formula? I only have a few attempts left so I didn't want to check if the answer was right.

 

[Doc Al]

Is this now the correct formula? I only have a few attempts left so I didn't want to check if the answer was right.

Yes, that's the correct formula. I haven't checked your calculation, but I'd be happy to do so later when I have some time. (If you can wait a few hours; I'll be off line for a bit.)

 

[jzwiep]

If you wouldn't mind doing so that'd be great. Thanks.

 

[Doc Al]

I get a different answer.

Etotal=Us=Up
1/2k(A+dL)^2=mg(A+dL+h)

So far, so good.

I rearrange with equilibrium forces equation to get:

72k^2-145mk-48m^2=0

Show how you got this one.

 

[jzwiep]

dL=mg/k

1/2k(A+mg/k)$$^{2}$$=mg(A+mg/k+h)

1/2kA$$^{2}$$ + 1/2m$$^{2}$$g$$^{2}$$/k = mg(A+h) + m$$^{2}$$g$$^{2}$$/k

1/2k$$^{2}$$A$$^{2}$$- kmg(A+h)-1/2m$$^{2}$$g$$^{2}$$=0

 

[Doc Al]

dL=mg/k

1/2k(A+mg/k)$$^{2}$$=mg(A+mg/k+h)

Looks good.

1/2kA$$^{2}$$ + 1/2m$$^{2}$$g$$^{2}$$/k = mg(A+h) + m$$^{2}$$g$$^{2}$$/k

Ah... looks like you made an error when you expanded (A+mg/k)².

(a + b)² ≠ a² + b²

 

[jzwiep]

Ah, that was silly of me

dL=mg/k

1/2k(A+mg/k)$$^{2}$$=mg(A+mg/k+h)

1/2kA$$^{2}$$ + Amg + 1/2m$$^{2}$$g$$^{2}$$/k = mg(A+h) + m$$^{2}$$g$$^{2}$$/k

1/2k$$^{2}$$A$$^{2}$$- kmgh-1/2m$$^{2}$$g$$^{2}$$=0

So changing that gives me:

72k$$^{2}$$ - 27.44mk - 48m$$^{2}$$

Which works out to:

k=1.02899m

Giving me a:

f=http://www.wolframalpha.com/input/?i=sqrt(1.02899)/(2pi)"

Is that correct?

 

[Doc Al]

I'd say you're off by a factor of 10, so double-check your arithmetic.

 

[jzwiep]

Okay it looks like I did the quadratic wrong,

(27.44m + 117.895m)/144

k=1.00927m

Which gives f=0.15989Hz

But that's not a factor of 10 off. I double checked my algebra for:

1/2k$$^{2}$$A$$^{2}$$ - kmgh-1/2m$$^{2}$$g$$^{2}$$ =0

And it looks right. Do you know if I'm getting my error from here? or from my quadratic?

 

[Doc Al]

But that's not a factor of 10 off. I double checked my algebra for:

1/2k$$^{2}$$A$$^{2}$$ - kmgh-1/2m$$^{2}$$g$$^{2}$$ =0

And it looks right. Do you know if I'm getting my error from here? or from my quadratic?

That equation looks fine. You're making an error when you solve the quadratic.

 

[jzwiep]

Alright, thanks for the help.