A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Find the time period in seconds. A ) 5 2 π B ) 4 π 5 C ) 2 π 3 D ) 5 π

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A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Find the time period in seconds.   A  )            5              2      π        B  )            4      π              5        C  )            2      π              3        D  )            5        π

Yaretzi Huynh · Accepted Answer

The right answer is B             4      π              5      The link between velocity and displacement is given by  v  =  ω            A      2        −          x      2       The equation for the acceleration-displacement relationship is      |    a      |    =      ω    2    x Given that,      |        v    →        |    =      |        a    →        |    ⇒  ω            A      2        −          x      2        =      ω    2    x  ⇒      ω    2    (      A    2    −      x    2    )  =      ω    4        x    2   We have that amplitude A=3 cm &amp;amp; required distance x=2 cm  ⇒      ω    2    =                    A        2            −              x        2                    x      2        =            9      −      4        4    =      5    4    ⇒  ω  =            5        2    ⇒  T  =            4      π              5       seconds  (  ∵  T  =            2      π        ω  )

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