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Homework Help: Simple Harmonic Motion problem 1

  1. Dec 13, 2012 #1
    Hi friends the problem is -


    Attempt -

    As per the problem states,

    For the first second equation of SHM, (using x = A sin ωt)

    a = A sin ω

    From here I get, sin ω = a/ A --------------(1)

    Using it for the 2 seconds, total displacement would be a + b. So,

    a + b = A sin 2ω

    i.e. a + b = 2A sin ω. cos ω

    i.e. a + b = 2A sin ω. √1-sin2ω ----------------(2)

    Solving both the equations, I am getting the result,

    A = 2a2/ √(3a2 - b2 - 2ab)

    But the correct answer of this problem is option (A) as per the question.

    Please friends help me in solving this Problem.

    Thank you all in advance.
  2. jcsd
  3. Dec 13, 2012 #2


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    Note that the particle starts from rest. Does the equation x = Asinωt describe a particle starting from rest? (Assuming "start" means t = 0)
  4. Dec 14, 2012 #3
    No, thaks. I have taken the eqn in wrong manner.

    Let the equation be x = A cosωt

    Solving, a = A cos ω

    i.e. cosω = a/A

    And (a + b) = A cos 2ω

    i.e. (a + b) = A √(2cos2ω -1)

    Soving both we get A = √(a2 - b2 -2ab)

    Yet answer is not achieved
  5. Dec 14, 2012 #4


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    The distance travelled can not be x=Acos(ωt), as the distance travelled would be A at zero time, (it should be zero) and the velocity would be negative...

    If x(t) is the position at time t, the velocity is v=dx/dt. The particle travels in the positive x direction, and it is velocity is zero at t=0. What should be the form of the displacement as function of t?

  6. Dec 14, 2012 #5
    Creating such kind of function is really giving me problem. Please help me out.
  7. Dec 14, 2012 #6


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    What about Δx=A(1-cosωt)?

  8. Dec 14, 2012 #7
    Yes this function is proper for the question. I appreciate your help. I will try to find answer with its help and get back to you soon.
  9. Dec 14, 2012 #8
    I am currently doing SHM, I am willing to know how you arrived at that equation. :)
  10. Dec 14, 2012 #9
    I think its answer.

    In my notion he thought that -
    We are not allowed to use x = A sin ωt or x = A cos ωt here because these functions are not giving the appropriate conditions according to question. But the function which can give correct answer for v = 0 at t = 0 is is x = A cos ωt. But it it is giving some displacement at t = 0.
    so what it giving the displacement at t = 0. That is A. Subtract it from A and the displacement as well as the velocity both will be zero at t = 0.

    Even I am not sure that the function is correct or not right now. I will solve it and then again will discuss about it.
  11. Dec 15, 2012 #10

    Please tell me that why are saying that the displacement of the particle should be zero at t = 0 ?

    If I stretch a spring mass system fro it mean position and hold it for some time and then release. So there the particle would have some displacement at t = 0. Isn't it ?
  12. Dec 15, 2012 #11
    Thank you very much ehild I go the correct answer. But Please clear the doubt which raised earlier.

    The function is absolutely correct.
  13. Dec 15, 2012 #12


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    Pranav: The shm along x in general is x=A cos (ωt+θ). The displacement from the position at zero time is Δx=x-x0=Acos(ωt+θ)-x0.

    The particle starts from rest, so v=-Aωsinθ=0. That means θ=0 or π.
    It starts to move in the positive x direction, v should increase, so the acceleration at t=0 is positive: -Aω2cosθ>0, that is θ=π.
    cosθ=-1, but the displacement is zero if x0=Acosθ=-A.
    So you can write up the displacement as Δx=Acos(ωt+π)+A=A(1-cosωt).

  14. Dec 15, 2012 #13


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    Displacement is the change of position Δx=x(t)-x(0). Of course, it is zero at t=0. It is not meant as the change of length of a spring here.

    We start measuring time when the particle has zero velocity. Assuming a general SHM x(t)=Acos(ωt+θ), we can get theta from the conditions given, that the initial velocity is zero, and the particle has positive velocity after. Read my previous post.

  15. Dec 15, 2012 #14
    Thanks ehild! :smile:
  16. Dec 15, 2012 #15
    Very nice explanation.
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