Spring energy problem

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  • #1
bcjochim07
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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 of 3 cm above the top of the spring. The block sticks to the top end of the spring and then oscillates with an amplitude of 20 cm. What is the oscillation frequency?


Homework Equations





The Attempt at a Solution



I tried to use conservation of energy, saying that yo = 0 and the top of the spring

so mg(.03m)= .5k(.10m)^2

k= 6mg then I plugged that into f= (1/2pi)* sqrt (k/m)
But that didn't work. I'm think that maybe since it says "sticks" this is an inelastic collision where energy isn't conserved.

Also if y=0 at the top of the spring, as it compresses to its maximum does this lead to a negative gravitational potential energy in addition to the potential energy in the spring?
 

Answers and Replies

  • #2
Doc Al
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I tried to use conservation of energy, saying that yo = 0 and the top of the spring

so mg(.03m)= .5k(.10m)^2
There's nothing wrong with using conservation of energy (in fact you must). Compare the initial energy (at 3 cm above the spring) to the energy when the spring is maximally compressed. Hint: Measure gravitational PE from the lowest position.
 
  • #3
bcjochim07
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So since it compresses .10 cm, is this what I should do:

mg(.13m)=.5k(.10cm)^2

k= 26 mg

frequency= 1/2pi * sqrt [k/m] frequency = 1/2pi * sqrt [26mg/m] = 1/2pi* sqrt 26g

frequency= 2.54 Hz... hmmm that's not right
 
  • #4
Doc Al
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So since it compresses .10 cm,
Rethink that. The amplitude is 20 cm. Where is the equilibrium point?
 
  • #5
bcjochim07
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I'm not sure I understand what you're saying. The problem says that the amplitude is 10 cm. Isn't the top of the spring the equilibrium point?
 
  • #6
Doc Al
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I'm not sure I understand what you're saying. The problem says that the amplitude is 10 cm.
In your first post you give the amplitude as 20 cm.
Isn't the top of the spring the equilibrium point?
No. The equilibrium point is where the net force on the mass is zero.
 
  • #7
bcjochim07
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I'm not seeing it...I'm sorry Ok yeah the amplitude should be 10 cm
 
  • #8
Doc Al
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The mass will oscillate about the system's equilibrium point. Find that point by finding where the net force on the mass is zero. (Only two forces act on the mass.)
 
  • #9
bcjochim07
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so where the kx=mg

that point in the compression is x=k/mg how do I fit this into my expression?
 
  • #10
bcjochim07
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.3+k/mg is y=0 ?
 
  • #11
bcjochim07
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Well now I tried kinematics and found the velocity of the block before it hits the spring to be .767 m/s. Then I used the amount of compression of the spring to be the distance over which the mass accelerates in the positive direction (slowing down)

v^2=0m/s -2(9.80)(.03)
v= .767m/s

0m/s=.767m/s+2a(.10)
a=-3.835

So the force on the block is F=m(3.835)

Fsp=kx k(.10)=(3.835)m
k=38.25m

use eqn freq= (1/2pi)*sqrt(k/m) freq= (1/2pi)* sqrt (38.25) = .984 ... not right either
 
  • #12
Doc Al
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so where the kx=mg

that point in the compression is x=k/mg how do I fit this into my expression?
That point, a distance x below the initial position of the spring, is the equilibrium point. You know the amplitude of the motion, so how far down does the mass compress the spring at the lowest point?
 
  • #13
bcjochim07
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Well, if the amplitude is .10m then the maximum compression is .10m below the equilibrium point. I think I'm having a problem visualizing this. So if it's dropped from 3cm above the top of the spring, then does it compress 10 cm to the equilibrium point and then 10 cm below that?
 
  • #14
Doc Al
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Well, if the amplitude is .10m then the maximum compression is .10m below the equilibrium point.
Good.
I think I'm having a problem visualizing this. So if it's dropped from 3cm above the top of the spring, then does it compress 10 cm to the equilibrium point
No. Realize that the distance from the original position of the top of the spring to the equilibrium point is not 10 cm. (It hits the spring with some speed.) You already figured out the distance it must compress to get to the equilibrium point in post #9; use it.
and then 10 cm below that?
Right.

(This is a tricky one to visualize; don't give up on it.)
 
  • #15
bcjochim07
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So is the initial height above the lowest point( where it is compressed the most) is .03m + k/mg + .10m ? If that's right then would I do this: (.13+ k/mg)mg = .5k(.10m)^2?
 
  • #16
Doc Al
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So is the initial height above the lowest point( where it is compressed the most) is .03m + k/mg + .10m ?
Good!
If that's right then would I do this: (.13+ k/mg)mg = .5k(.10m)^2?
Almost. How much is the spring compressed? (Not just .10 m.)
 

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