Spring Passing Equilibrium Position

  • #1
http://i662.photobucket.com/albums/uu347/TwinGemini14/showme2.gif

A box of mass m slides at an initial speed v into a relaxed spring of spring constant k.

How long is the box in contact with the spring before it passes the equilibrium position again?

A) tcontact = 2p [k/m]½
B) tcontact = p [m/k]½
C) tcontact = ½ p [m/k]½

p here represents pi.

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I believe the correct answer is A, because the box is with contact will the spring for a full oscillation. So 2pi.

Can somebody help me on this one, I seem to be really stuck. Thanks in advance!

(I posted a link to a picture of the system on the top.)
 

Answers and Replies

  • #2
Doc Al
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I believe the correct answer is A, because the box is with contact will the spring for a full oscillation.
Why do you think it's in contact for a full oscillation? What would represent a full oscillation?
 
  • #3
I just assumed that when the block hits the spring, it compresses it and then the spring restores back to its equilibrium position, thus releasing the block then. In that case, it travels one complete oscillation. That was my assumption. Can somebody please explain the physics behind this problem?
 
  • #4
Doc Al
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I just assumed that when the block hits the spring, it compresses it and then the spring restores back to its equilibrium position, thus releasing the block then. In that case, it travels one complete oscillation.
When a mass at the end of spring oscillates, it goes from one extreme to the other on each side of the equilibrium point. Here you start at equilibrium, go to full compression (one extreme), then pass the equilibrium point. What would be next for a full oscillation?
 
  • #5
Oh, so wouldn't it just be pi?
 
  • #6
Doc Al
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Oh, so wouldn't it just be pi?
The box is in contact with the spring for half the period of a full oscillation.
 

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