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Math Help - Simple Harmonic Motion 2

  1. #1
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    Exclamation Simple Harmonic Motion 2

    A particle in SHM passes through two points A and B, 22 cm apart, with the same speed, having spent 2 seconds in passing from A to B. After another 2 seconds, it returns to B. Find the period and the amplitude of the motion.
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  2. #2
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    Quote Originally Posted by UltraGirl View Post
    A particle in SHM passes through two points A and B, 22 cm apart, with the same speed, having spent 2 seconds in passing from A to B. After another 2 seconds, it returns to B. Find the period and the amplitude of the motion.
    ignore this , see next .
    Last edited by mathaddict; May 8th 2010 at 08:04 AM.
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  3. #3
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    Quote Originally Posted by UltraGirl View Post
    A particle in SHM passes through two points A and B, 22 cm apart, with the same speed, having spent 2 seconds in passing from A to B. After another 2 seconds, it returns to B. Find the period and the amplitude of the motion.
    Since the particle in SHM passes through two points A and B, 22 cm apart, with the same speed, they must be in the opposite sides of the equilibrium. It takes one second to reach B from the equilibrium position. The particle moves further, reaches the maximum amplitude and returns back to B in 2 seconds. So the particle takes 2 seconds to reach the maximum position from the equilibrium position.
    Hence the period of SHM is 8 seconds.
    Equation of SHM is
    A = Aosin(ωt)
    In one second the amplitude is 11 cm. So
    11 = Aosin(2π/8)
    Ao = 11*sqrt(2)
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