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Homework Help: SHM question with pulley and springs

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the time period of the oscillations of the body whose mass is m when it is displaced slightly form equilibrium. All springs have spring constant k and the pulleys and strings are massless and frictionless.

    2. Relevant equations

    where T is the time period of oscillations
    m is the mass of the body, K is the effective force constant.
    3. The attempt at a solution

    I've tried to find the net displacement hoping that it would help.
    When the body is pulled by a distance x from equilibrium, the spring 1 extends by x. Then this force due to extension on the pulley must be balanced by the tension and extension of spring 2. Since the latter two are equal, spring 2 must extend by 0.5x. Similarly, spring 3 must extend by 0.25x.

    So the next displacement of the block is [itex]\frac{7x}{4}[/itex].

    I don't know what to do after that to find the effective force constant on the block.
  2. jcsd
  3. Apr 4, 2013 #2
    Consider just one spring and pulley connected as shown. What is its equivalent spring?
  4. Apr 4, 2013 #3
    Well since the total displacement is 7x/4 and if there were a spring that would extend by x and still provide the same force as the one connected to the mass, would the force constant be 7k/4 ?

    I don't exactly know how the force constant is to be calculated. Could you please give hints in that direction?

    Thanks a lot.
  5. Apr 4, 2013 #4
    Forget total displacement. Just consider one pulley and one spring. If the spring's constant is k, and the pulley's displacement is x, what is the resultant force exerted by the pulley on the object attached to it?
  6. Apr 4, 2013 #5
    Ok the force is kx. What next?
  7. Apr 4, 2013 #6
    That would mean you can just remove all the pulleys out of the picture. Does that look plausible?

    How did you get the kx answer to begin with? If the pulley's displacement is x, what is the displacement of the spring's end?
  8. Apr 4, 2013 #7
    If the pulley at the bottom is displace by x, the block at the bottom must be displaced by 7x/2 right? Because for the forces on the pulley to balance, upward force (2kx) must be balanced by downward force. And thus spring 1 extends by 2x. Similarly spring 3 will be extending by x/2 to balance forces on the top pulley.

    Is that correct?
    Last edited: Apr 4, 2013
  9. Apr 4, 2013 #8
    You keep talking about the block at the bottom and the balance. Don't. Just consider the pulley and the spring as if nothing else was there. You have already found that when the pulley's displacement is x, the spring extension is 2x. Now, if the spring exerts force F, what force does the pulley exert to a mass attached to its center?
  10. Apr 4, 2013 #9
    Could you tell me which spring I am to consider. If it is the bottom spring and is exerting a force F due to extension of 2x on the pulley, then It must exert a force F in the upward direction on the mass attached right?
  11. Apr 5, 2013 #10
    Consider the top left spring-pulley subsystem. Imagine that the rest of the system is removed. Imagine that you pull the pulley at its middle point down so that its displacement is x. Find the force it is exerting on you in this position.
  12. Apr 5, 2013 #11
    Ok then the force on me because of the pulley is 2kx.
  13. Apr 5, 2013 #12
    Explain why.
  14. Apr 5, 2013 #13
    The force on the pulley upwards is kx + the tension in the wire. But these two forces have to be equal as the string is massless. So net upward force on the pulley is 2kx and thus I should be pulling it downwards with an equal magnitude because the pulley is massless. So by newton's 3rd law it should be pulling me up with force 2kx. Is that correct?

  15. Apr 5, 2013 #14
    It is correct that if the spring is stretched so that its reaction force is F, the pulley will produce 2F.

    However, if the pulley's displacement is x, what is the spring's end displacement? So what is that F above?
  16. Apr 5, 2013 #15
    The spring displacement will be x/2? So is the pulley pulling me upwards with F=kx itself ? Because the net upward force on the pulley is kx?
  17. Apr 5, 2013 #16
    Why is it x/2? Draw a picture. The pulley goes down by x. That means the length of the rope to its left and the length of the rope to its right must increase by x EACH. So what is the total increase of the length?
  18. Apr 5, 2013 #17
    OK. The net increase in length is 2x. SO the spring force is 2kx and the tension is also 2kx. So the net force on me is 4kx?
  19. Apr 5, 2013 #18
    Correct. So you have replaced the spring-pulley system with just a spring whose coefficient is 4 times the coefficient of the original spring. Now in the top left you have two springs in series, 4k and k. What is the equivalent spring? When that is done, you will again have a spring-pulley system, and you know how to deal with it.
  20. Apr 5, 2013 #19
    I hope you will be able to finish that. I have to go catch a plane. I will check back tomorrow. Good luck!
  21. Apr 5, 2013 #20
    Right. The equivalent spring constant of the series combination of the two springs is 4k/5. So by the previous argument, I should replace it with a spring of four times the spring constant (16k/5) and now I have two springs again in series (k and 16k/5) and their effective spring constant is 16k/21 right?

    So the Time period of oscillation is [itex]2∏\sqrt{\frac{21m}{16k}}[/itex].

    Thanks a lot voko. I couldn't have done it without you.:biggrin:
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