Effect of dropping putty on a block doing SHM

  • #1
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Hi

here's the problem i need help with.

a block is attached to a spring and performs SHM along a horizontal straight line. a piece of
putty is dropped vertically so that it sticks to the block when it lands on it. refer to the attached figure.
Discuss qualitatively the effects of this, if any, on the amplitude and the period of oscillation in the following two separate cases in which the putty lands on the block.
a) when the block moves past its equilibrium position
b) when the block is momentarily at rest at maximum displacement.

now here's what i think. since the putty sticks to the block, its an inelastic collision.
so the energy is not conserved. energy is lost in heat. now E=(1/2)k A^2 when the block is at maximum displacement, where A is the amplitude. since spring is not changed, k will not change, so for E to decrease, A must decrease. so as a result amplitude will decrease.
is this correct ? also i am not able to say what will happen to the period and what would be the difference between questions a) and b).

please help
Issace n

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Homework Equations





The Attempt at a Solution

 

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Answers and Replies

  • #2
kuruman
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Do you have an expression for the period (or frequency) of a spring-mass system? What happens to this expression if the mass is made larger?

You are correct in viewing this as an inelastic collision that does not conserve mechanical energy. However, the energy stored in the oscillator is not the only mechanical energy there is. The putty also has potential energy initially which is converted to kinetic energy just before it hits. So while the total initial mechanical energy of oscillator plus putty decreases in this inelastic collision, the oscillator's mechanical energy depends only on the amplitude of the oscillator. Does the amplitude of this oscillator change after the collision?
 
  • #3
912
19
Do you have an expression for the period (or frequency) of a spring-mass system? What happens to this expression if the mass is made larger?

You are correct in viewing this as an inelastic collision that does not conserve mechanical energy. However, the energy stored in the oscillator is not the only mechanical energy there is. The putty also has potential energy initially which is converted to kinetic energy just before it hits. So while the total initial mechanical energy of oscillator plus putty decreases in this inelastic collision, the oscillator's mechanical energy depends only on the amplitude of the oscillator. Does the amplitude of this oscillator change after the collision?

sorry for the late response. was out of town. now
w^2 = k/m. So T= 2*pi / w ---> T= (2*pi) sqrt(m/k). so I see that if mass increases, period should increase.
since putty also has kinetic energy at the moment of hitting the block, we don't know if the energy of the system (block+putty) will increase or decrease even though its an inelastic collision, right ?
also how do we distinguish between case a) and case b).

Regards
Issac
 
  • #4
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please help me on this problem
 
  • #5
kuruman
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You are correct, sir Isaac, about the change in frequency. For the other qualitative question, use momentum conservation to consider the velocity change of the oscillator from immediately before to immediately after the putty sticks. The key question is, does the collision add or subtract from the initial energy of the oscillator? Note that in case (a) the block may be moving up or down past the equilibrium position, while in case (b) the oscillator is instantaneously at rest when the collision occurs.
 

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