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Math Help - Simple harmonic motion period twice as book's answer :help:

  1. #1
    Member ssadi's Avatar
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    Simple harmonic motion period twice as book's answer :help:

    A light string has its two ends fastened together to form a loop of natural length "2l" and modulus of elasticity mg. The string is placed over two smooth fixed pegs A and B where AB is horizontal and of length 2l. A particle P of mass 2m is attached to the mid-point of the lower part of the loop and hangs in equilibrium vertically below C, the mid-point of AB.

    d)Find the period of the motion.

    Code:
    a) Explain why the extension of the string is equal to twice the length of AP.
    b) Show that the distance CP is l.
    P is projected vertically downwards from its equilibrium position.
    c) Show that, providing the string does not break and P does not reach the level of AB, the subsequent motion is simple harmonic.
    a), b), c) worked out but my period {d)} is 4\pi \sqrt \frac{l}{g} while book's answer is 2\pi \sqrt \frac{l}{g}. I took, in the formula, mass as 2m and original length as 2l, as given question, but the answer disagrees. Anybody with a suggestion why this might be? :S :help:
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  2. #2
    Member TheMasterMind's Avatar
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    Quote Originally Posted by ssadi View Post
    A light string has its two ends fastened together to form a loop of natural length "2l" and modulus of elasticity mg. The string is placed over two smooth fixed pegs A and B where AB is horizontal and of length 2l. A particle P of mass 2m is attached to the mid-point of the lower part of the loop and hangs in equilibrium vertically below C, the mid-point of AB.

    d)Find the period of the motion.

    Code:
    a) Explain why the extension of the string is equal to twice the length of AP.
    b) Show that the distance CP is l.
    P is projected vertically downwards from its equilibrium position.
    c) Show that, providing the string does not break and P does not reach the level of AB, the subsequent motion is simple harmonic.
    a), b), c) worked out but my period {d)} is 4\pi \sqrt \frac{l}{g} while book's answer is 2\pi \sqrt \frac{l}{g}. I took, in the formula, mass as 2m and original length as 2l, as given question, but the answer disagrees. Anybody with a suggestion why this might be? :S :help:

    a)
    consider forces along the strings and the weight of the particle.resolve the string forces along weight of the particle,something along the lines of
    2mgcosq=mg
    would be an appropriate equation

    b)
    you can resolve forces on the particle in terms of some angle q. Then use the relation of part (a) to find an equation for the tension in the string. You also need a trigonometric relation between x=AP and l. These three equations can be solved to work out q.

    c)hint: show the acceleration is proportional to the displacement


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  3. #3
    Member ssadi's Avatar
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    Quote Originally Posted by TheMasterMind View Post

    a)
    consider forces along the strings and the weight of the particle.resolve the string forces along weight of the particle,something along the lines of
    2mgcosq=mg
    would be an appropriate equation

    b)
    you can resolve forces on the particle in terms of some angle q. Then use the relation of part (a) to find an equation for the tension in the string. You also need a trigonometric relation between x=AP and l. These three equations can be solved to work out q.

    c)hint: show the acceleration is proportional to the displacement


    I did managed a), b) and c), I was facing problem in "d)", the period of motion. Your effort went futile here
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