A piece of clay stuck to a SHM oscillator

In summary, the total mechanical energy is not conserved in this situation due to the loss of mechanical energy of the falling clay. Additionally, the vertical momentum of the system is not conserved, but the horizontal momentum is approximately conserved due to the impulse of the spring force during the impact time interval. However, it is not specified if the question is referring to the horizontal momentum of the block or of the block+clay system.
  • #1
Leo Liu
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Homework Statement
A quiz question
Relevant Equations
##E_i=E_f##
1605496384146.png

I don't get why the total mechanical energy is not conserved in this situation.
When the length of the spring reaches the maximum, the speed of the block is 0 and we have the following equation:
$$E=K+U=1/2mv^2+1/2kA^2,\text{where A is the amplitude} \implies E=1/2kx_{max}^2$$
I can't see why the amount of energy changes. Could someone please explain the reason to me? Thanks a lot.
 
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  • #2
Leo Liu said:
I can't see why the energy changes.
Read it more carefully. It says the ME of the falling clay is lost.
 
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  • #3
haruspex said:
Read it more carefully. It says the ME of the falling clay is lost.
Aha, I see--this is kinda tricky. Thank you.
 
  • #4
If we want to extent this a bit, i think the vertical momentum of the system is not conserved either.
 
  • #5
Delta2 said:
If we want to extent this a bit, i think the vertical momentum of the system is not conserved either.
Well, it is conserved if you consider the movement of the Earth :P
 
  • #6
Leo Liu said:
Well, it is conserved if you consider the movement of the Earth :P
Well, yes if we include Earth into our system it is conserved but by system i meant the block+clay.
 
  • #7
Leo Liu said:
Homework Statement:: A quiz question
Relevant Equations:: ##E_i=E_f##

I don't get why the total mechanical energy is not conserved in this situation.
When the length of the spring reaches the maximum, the speed of the block is 0 and we have the following equation:

I'm with you. No mention is made of the height of the drop...it could be made negligibly small in which case the problem is better. Badly stated problem by my reckoning.
 
  • #8
Delta2 said:
If we want to extent this a bit, i think the vertical momentum of the system is not conserved either.
Right, but it explicitly asks about horizontal momentum.
 
  • #9
haruspex said:
Right, but it explicitly asks about horizontal momentum.
I have my doubt about conservation of horizontal momentum because we have the force of the spring acting in the x- direction which is at a maximum value btw during the collision.
I think "none of these" is the ultimate correct answer after all but of course one may argue that the horizontal momentum is approximately conserved.
 
  • #10
Delta2 said:
I have my doubt about conservation of horizontal momentum because we have the force of the spring acting in the x- direction which is at a maximum value btw during the collision.
I think "none of these" is the ultimate correct answer after all but of course one may argue that the horizontal momentum is approximately conserved.
I don't see how it affects horizontal momentum. The block is stationary when the clay sticks to it. No horizontal momentum immediately before impact and none immediately after.
 
  • #11
haruspex said:
and none immediately after
Isn't the spring force , via its impulse, going to transfer horizontal momentum ##\int_t^{t+\Delta t} kx(t) dt## during the impact time interval ##(t,t+\Delta t)##.
 
  • #12
Delta2 said:
Isn't the spring force , via its impulse, going to transfer horizontal momentum ##\int_t^{t+\Delta t} kx(t) dt## during the impact time interval ##(t,t+\Delta t)##.
Yes, but in all such questions we take Δt to be arbitrarily small.
 
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  • #13
Btw, the question could be clearer. It doesn’t specify whether it's the horizontal momentum of the block or of the block+clay system. If the latter, it is unnecessary to assume the block is stationary for the duration of the impact.
 
  • #14
haruspex said:
Btw, the question could be clearer. It doesn’t specify whether it's the horizontal momentum of the block or of the block+clay system. If the latter, it is unnecessary to assume the block is stationary for the duration of the impact.
Judging by the answer i think it means the horizontal momentum of the clay.
 
  • #15
Delta2 said:
Judging by the answer i think it means the horizontal momentum of the clay.
The answer given is that neither the clay nor the block change in momentum.
Given the question but not the answer, I would have guessed it was asking about the block.
 

1. What is SHM and how does it relate to the clay stuck to the oscillator?

SHM stands for Simple Harmonic Motion, which is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium. In this case, the clay stuck to the oscillator is experiencing SHM because the force of the oscillator is causing it to move back and forth in a periodic manner.

2. Why does the clay stick to the oscillator in the first place?

The clay sticks to the oscillator due to the force of adhesion, which is the attractive force between two different materials. In this case, the clay and the oscillator have different materials and therefore experience adhesion, causing the clay to stick.

3. Can the clay affect the oscillator's motion?

Yes, the clay can affect the oscillator's motion. Since the clay is stuck to the oscillator, it adds an additional mass to the system, which can change the oscillator's natural frequency and amplitude. This can alter the overall motion of the oscillator.

4. How does the clay's position on the oscillator affect its motion?

The clay's position on the oscillator can affect its motion in a few ways. If the clay is placed closer to the center of the oscillator, it will experience a smaller amplitude and a higher frequency of motion. If the clay is placed closer to the edge of the oscillator, it will experience a larger amplitude and a lower frequency of motion.

5. Can the clay eventually fall off the oscillator due to its motion?

Yes, the clay can eventually fall off the oscillator due to its motion. If the amplitude of the oscillator's motion is large enough, it can cause the clay to lose its adhesion and fall off. Additionally, if the oscillator's motion is too fast, the clay may not have enough time to re-adhere to the oscillator before it moves again, causing it to fall off.

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