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A particle performing linear S.H.M. has ...

A particle performing linear S.H.M. has a maximum displacement of 0.1 m. Its acceleration at a distance of 0.03 m from its position is 0.12 `m//s^2`. What is its velocity at a distance of 0.06 m from its mean position ?

Text Solution

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`v=omegasqrt(a^2-x^2) = sqrt(ac c^n/x)xsqrt(a^2-x^2)`
`sqrt0.12/0.13xxsqrt((0.1)^2-(0.06)^2)`
`2sqrt(0.01-0.0036)`
`2sqrt0.0064=2xx0.08=0.16m/s.`
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Knowledge Check

  • A particle performings S.H.M. with amplitude 'a' has maximum velocity v, its speed at displacement a//2 will be

    A
    2 v
    B
    0.866 v
    C
    0.5 v
    D
    0.44 v

  • The velocity of a particle performing S.H.M. at mean position is,

    A
    Maximum
    B
    Gradually increases
    C
    Minimum
    D
    Gradually decreases

  • The maximum displacement of a particle performing S.H.M. from its mean position is

    A
    amplitude of S.H.M.
    B
    frequency of S.H.M.
    C
    period of S.H.M.
    D
    wavelength of S.H.M.

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