Return time for a string

  • Thread starter aurao2003
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Homework Statement


Hi

I am struggling with the second part of this problem. It reads this way:



A particle P of mass 0.8kg is attached to one end of a light elastic string of natural length 1.6m and modulus of elasticity 20N. The other end of the string is fixed to a point O on the smooth horizontal surface on which P rests. The particle is held at rest with OP = 2.6m and then released.

a)Show that the, while the string is taut, P moves with simple harmonic motion.

b)Calculate the time from the instant of release until P returns to its starting point for the first time?



How do I start the b part? Please help!!!




Homework Equations


Hookes Law



The Attempt at a Solution

 

Answers and Replies

  • #2
311
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Well, if you were able to do the first bit, then you showed that the equation of motion of P is of the form

[tex] m \ddot{x} = -\omega^2 x [/tex]

The time taken for the particle to come back to its starting point after one cycle is nothing but the time period of this oscillation.
 
  • #3
126
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Well, if you were able to do the first bit, then you showed that the equation of motion of P is of the form

[tex] m \ddot{x} = -\omega^2 x [/tex]

The time taken for the particle to come back to its starting point after one cycle is nothing but the time period of this oscillation.
Hmm! Will I not need to take into consideration the time it took the string to slacken?
 
  • #4
311
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Ah! good point, I missed that completely. Sorry.

Yes, you have to take that into consideration. I'll describe the motion to you, and the further calculation should be easy.

Lets put the origin of our coordinate system at O and the motion is along the x-axis.

The particle starts out at P (2.6,0). Till (1.6,0) the motion is simple harmonic (how much time does that take?). The particle then moves from (1.6,0) to (-1.6,0) with uniform velocity (how much?). Again it undergoes SHM from (-1.6,0) to (-2.6,0) and back to (-1.6,0). Yet again, it moves uniformly from (-1.6,0) to (1.6,0). And again undergoes SHM from (1.6,0) to (2.6,0). And its back after one round!

Can you now calculate the time?
 
  • #5
126
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Ah! good point, I missed that completely. Sorry.

Yes, you have to take that into consideration. I'll describe the motion to you, and the further calculation should be easy.

Lets put the origin of our coordinate system at O and the motion is along the x-axis.

The particle starts out at P (2.6,0). Till (1.6,0) the motion is simple harmonic (how much time does that take?). The particle then moves from (1.6,0) to (-1.6,0) with uniform velocity (how much?). Again it undergoes SHM from (-1.6,0) to (-2.6,0) and back to (-1.6,0). Yet again, it moves uniformly from (-1.6,0) to (1.6,0). And again undergoes SHM from (1.6,0) to (2.6,0). And its back after one round!

Can you now calculate the time?
Sorry for the delay of getting back to you. Its the scourge of the full time worker! I will give you a calculated response tonight. Thanks.
 

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