More SHM problems

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  • #1
bcjochim07
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Homework Statement


#1 A spring with spring constant k is suspended vertically from a support and a mass m is attached. The mass is held at the point where the string is not streteched. Then the mass is released and begins to oscillate. The lowest point in the oscillation is 20 cm below the point where the mass was released. What is the oscillation frequency?

#2 A 300g oscillator has a speed of 95.4 cm/s when its displacement is 3 cm and a speed of 71.4 cm/s when its displacement is 6 cm. What is the oscillator's maximum speed.

Homework Equations





The Attempt at a Solution


#1 shouldn't be that difficult of a problem, but it's giving me trouble
Here's what I know: amplitude = .20 m
The total energy= 1/2k(.20 m)^2= 1/2m(vmax)^2
vmax= omega*A

But I can't figure out how to substitute the equations into each other to get the correct answer, which is 1.58 Hz

#2 I have also tried substituting the equations into each other:
3cm= Acos(omega*t+phi)
95.4 cm/s= -omega*A sin (omega*t+phi)

6 cm= Acos(omega*t + phi)
71.4 cm/s = -omega*A sin( omega*t + phi)

Any hints would be greatly appreciated. Thanks.
 

Answers and Replies

  • #2
bcjochim07
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any help would be appreciated
 
  • #3
alphysicist
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bcjoshim07,

problem #1. The amplitude in the first part is not 20cm. Be careful to think about how the mass was released.

problem #2. Try using energy conservation for the three points.
 
  • #4
bcjochim07
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i don't see how it wouldn't have an amplitude of 20 in #1 because it's released from equilibrium and isn't amplitude the maximum displacement from equilibrium?
 
  • #5
bcjochim07
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problem 2 I understand now. Thanks!
 
  • #6
alphysicist
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i don't see how it wouldn't have an amplitude of 20 in #1 because it's released from equilibrium and isn't amplitude the maximum displacement from equilibrium?

bcjochim07,

It was not released from equilibrium in this problem. Equilibrium is that point at which the spring is stretched enough so that the spring force is cancelling out the weight force. But this mass is released from the point where the spring is not stretched. What would that point be?
 
  • #7
bcjochim07
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where kx=mg but I'm still not sure where to go with it
 
  • #8
bcjochim07
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no wait that's not right
 
  • #9
bcjochim07
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hmmm... a point where the string is not stretched...I'm still not sure couldn't it be any point above where it would be if it were just hanging from the string?
 
  • #10
alphysicist
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So they release the mass from rest, and it goes downward a total of 20cm. How far up does it go after that?

How is the amplitude related to the total displacement that the mass undergoes while it is moving downwards from the highest to the lowest point?
 
  • #11
bcjochim07
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it is half of that amount so then the amplitude is 10cm? and then where should I go with that?
 
  • #12
alphysicist
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Write a force equation for the equilibrium point. (Where is the equilibrium point?) That will be an expression involving k and m. What do you get?

Then the normal formula for frequncy

[tex]
f=\frac{1}{2\pi} \sqrt{\frac{k}{m} }
[/tex]
 
  • #13
bcjochim07
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kx=mg

so k=mg/x but how do I know what the equilibrium point is?
 
  • #14
bcjochim07
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so equilibrium is at x=mg/k
 
  • #15
alphysicist
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Looks good; do you know what x is? Once you have that you're almost done.
 
  • #16
bcjochim07
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Ahhh... so the equilbrium point is at 10cm so k=mg/.1m

substituted that into the formula and get frequency= 1.58 Hz

I'm still a little unsure of how we defined our coordinates--we must have defined them as the lowest point being x=0
 
  • #17
alphysicist
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Well, we know the release point is at the positive amplitude, so I would call that x=+10 for purposes of the oscillation (using the sine or cosine function, etc.). (That's because the oscillation includes the effects of gravity.)

However, for purposes of calculating the total spring force, we would probably want to call x=0 that point at which the spring is unstretched. Because if it's not stretched it is not putting a force on the mass, so F=0.
 

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