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Changing the mass of a simple harmonic oscillator

  1. Jun 11, 2014 #1
    1. The problem statement, all variables and given/known data
    image.png

    2. Relevant equations

    3. The attempt at a solution
    I find this task very hard to understand. First of all, when adding more mass, wouldn't that change the acceleration, according to F=ma? And in that case the velocity should also change when adding more mass, shouldn't it? That would then also change the amplitude and angular frequency. When I think of it intuitively, I think that adding more mass will increase the inertia, and hence make the amplitude less and also the frequency. Why is this wrong?

    I don't understand how to think about the physical concepts in this example. If you understand it, please explain how you reason about it!
     
  2. jcsd
  3. Jun 11, 2014 #2

    TSny

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    When the second box is "attached to" the first box, you can think of it as a "collision" between the masses. Can you think of a physical quantity that is conserved during the "collision"?
     
  4. Jun 12, 2014 #3
    Hm first of all I don't see what makes the second box stick together with the first box, but maybe that doesn't matter. So let's think of it as a collision. Is it then correct to see this situation as equivalent to the second box was put in front of the first box, and then the first box collides with the second box and "drags with" the second box? In that case we have a completely inelastic collision, and kinetic energy is preserved.

    KEf = KE0, that is

    m1*vf1^2 + m2*vf2^2 = m1*v01^2 + m2*v02^2

    The initial velocity of the mass on the spring (m1) is vmax and for the mass being attached (m2) is 0, and after the collision the two masses will have the same velocity. So we can simplify to:

    (m1+m2)*vf^2 = m1*vmax^2

    m1=m2=m so we get 2*m*vf^2 = m*vmax^2 <=> 2*vf^2 = vmax^2 <=> vf = √2*vmax.

    This is unfortunately not correct. What is wrong in my reasoning?
     
  5. Jun 12, 2014 #4

    TSny

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    Yes, good. You can think of it as a completely inelastic collision.

    What is meant by the term inelastic? Is KE conserved in such a collision? If not, what is conserved?
     
  6. Jun 12, 2014 #5
    In any collision, nothing but the total energy and linear momentum is conserved. Check your calculations accordingly.
    Mass doubles, Energy is conserved, Find out what decreases and what increases from
    F=ma. or F=m.(Second derivative of position wrt time).... you have the additional relations to assist you, which are
    F=-kx and a=-(w^2).x and w(angular freq)=root(k/m) if you require them. :) all this keeping in mind the relationship between force and work(Total Energy) in a conservative process..... I'm sure you can solve this problem, this isn't the toughest.. :)
     
  7. Jun 12, 2014 #6

    TSny

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    Hello, SohamSen. Welcome to PF!

    You are right that total energy is conserved. But total energy includes all forms of energy, such as "heat" stored internally within the objects. Maxo will need to decide if KE alone is conserved in this type of collision.
     
  8. Jun 12, 2014 #7
    Right. The linear momentum is conserved. The equations that I did for KE will almost be the same for linear momentum:

    pf = p0

    m1*vf + m2*vf2 = m1*v01 + m2*v02

    The initial velocity of the mass on the spring (m1) is vmax and for the mass being attached (m2) is 0, and after the collision the two masses will have the same velocity. So we can simplify to:

    (m1+m2)*vf = m1*vmax

    m1=m2=m so we get 2*m*vf = m*vmax <=> 2*vf = vmax <=> vf = 1/2*vmax.

    But it's still not correct.

    I also still don't know what makes the second box stick to the first one? Maybe that doesn't matter, but why not? I also wonder, if we see this as an inelastic collision, is it then correct to see this situation as equivalent to a situation where the second box was put in front of the first box, and then the first box collides with the second box and "drags with" the second box? Please help to clarify so I can understand this.
     
    Last edited: Jun 12, 2014
  9. Jun 12, 2014 #8

    TSny

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    That all looks right to me. The wording of part (a) is a little vague. They ask "what happens" to vmax. I think you are right that it is reduced by a factor of 1/2. How do you know it's not the right answer?

    Unfortunately, the problem statement is not very clear on the meaning of "attached to". But I don't see any other way to interpret the problem except to assume that it's equivalent to a totally inelastic collision.

    Yes, that's right.
     
  10. Jun 12, 2014 #9
    In my book the answer is given as that the speed remains the same. It seems they assume that the second box attached will already have an initial speed equal to the first box. But it's not said clear in the task. I'm glad I'm not the only one who interpreted it this way anyway. :)

    What about the other questions, b) and c)? In b) they ask if the amplitude changes. I find this quite hard to solve also. I know the force of the spring relates to the amplitude like F=-k*A. Does this help finding out how the amplitude will change? How can we find that out?
     
  11. Jun 12, 2014 #10

    TSny

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    OK. With that interpretation, then the diagram in the text makes more sense to me now. Sorry for misleading you.

    You might want to consider energy here. Can you describe what type of energy the system has at x = 0? At x = Anew? Are these two energies related? Can you use this to relate Anew to vmax? How is Aold related to vmax?
     
  12. Jun 12, 2014 #11
    At x=0 the system has kinetic energy only. At Anew it is potential elastic energy. Yes they are related somehow. I know they can't both be max at the same time. So when PE=max KE=0 and vice versa. I also know that if there is no non conservative forces doing work, then KEf+PEf = KE0 + PE0. Here PE0=0 and KEf=0, so we get PEf = KE0 or

    1/2*k*A^2 = 1/2*m*vmax^2

    Am I on the right track? How can we continue from here?
     
  13. Jun 12, 2014 #12

    TSny

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    Yes, you're on the right track!

    But, be careful with the interpretation of the mass m. Apply this equation to two cases: (i) before the second mass is added (ii) after the second mass is added.
     
  14. Jun 12, 2014 #13
    Ah, I see now. Anew = √2*Aold. Great.

    Now I have another question, which I would also very much like explained. What would happen (concerning maximum speed, amplitude and angular frequency) if the second box was attached at the one extreme end of the oscillation cycle?

    I don't see how that would be different from when the second box is attached in the middle position, but apparently it is different. Can someone please explain why it is different?
     
  15. Jun 12, 2014 #14

    TSny

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    If you add the second mass to the first mass when x = A, have you added any energy to the system?

    What about when you add the second mass at x = 0?
     
  16. Jun 12, 2014 #15
    Well, the mass is increased but since the velocity at this point is 0 I guess there can't be any kinetic energy involved? Also the linear momentum must be zero for the same reason. And the elastic potential energy is unchanged since there is no displacement of the spring when the second box is added.

    Can this really be correct? According to my book the maximum velocity is actually decreased when the second box is added at one extreme end. But I don't understand why? How can we get to this conclusion?

    Then kinetic energy is added (like in the previous task).
     
  17. Jun 12, 2014 #16

    TSny

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    Yes, that's right. Adding the second mass to the first mass at x = A does not change the energy of the system.

    The energy of the system does not change when adding the second box at x = A.
    What does that tell you about the KE at x = 0? Is the total KE at x = 0 for the two boxes the same as it was for the one box at x = 0?

    Think about how the mass of the system affects the relationship between KE and speed.
     
  18. Jun 13, 2014 #17
    I think I understand. The only thing that matters it what happens to the total energy (TE) at the instant of changing some variable (in this case doubling mass when v=0). In this case the TE is unchanged because the box being added has no energy to add to the system. And since the TE is unchanged at this point, it's also unchanged at the point x=0, even though the KE is larger at x=0 compared to at x=A. Is that correct? But how does that go together with the fact that this is again equivalent to a inellastic collision, which by definition means KE is NOT preserved?

    I'm then wondering what happens to the PE(elastic), which leads us to the second question, which is what happens to the amplitude. In the first task the doubling of maximum KE also meant doubling of maximum PE. In this case the KE is unchanged, so the same reasoning should mean that PE must here also be unchanged. PE relates to the amplitude as PE=1/2*k*A^2. So if PE is the same, and k is the same, then A must also be the same? Again this is incorrect though. How can we solve this?
     
    Last edited: Jun 13, 2014
  19. Jun 13, 2014 #18

    TSny

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    Yes, that is correct.

    There is no collision. When the second box is added to the first box at x = A, both boxes are at rest. There is no loss of KE. Both boxes have zero KE just before they are attached and also zero KE just after they are attached. The energy of the simple harmonic motion does not change when the second box is added.

    In the first case where the second box was added at x = 0, there was also no collision. The second box had the same speed as the first box when the second box was added. Nevertheless, in this case energy was added to the simple harmonic motion because the second mass already had KE when it was added to the simple harmonic motion.

    I agree with you that for the case where the second box is added to the first box at x = A, the amplitude should not change. As you say, there is no change in total energy in this case. So the PE at x = A doesn’t change. Therefore the amplitude doesn’t change. I don’t understand why this isn’t the correct answer.

    [In the first case where the second box was added at x = 0, there is a change in amplitude.]
     
    Last edited: Jun 13, 2014
  20. Jun 13, 2014 #19
    Sorry I made a misstake, it actually does say the amplitude doesn't change. I mixed it up, it's the angular frequency that changes.

    I think I understand everything now. For some reason this physical example was one of the most unintuitive I've encountered in all of mechanics. But now it makes much more sense. Thanks TSny for clarifying so much, you've been very helpful. :)
     
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