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A point particle of mass 0.1kg is execut...

A point particle of mass `0.1kg` is executing `SHM` with amplitude of `0.1m`. When the particle passes through the mean position. Its `K.E.` is `8xx10^(-3)J`. Obtain the equation of motion of this particle if the initial phase of oscillation is `45^(@)`.

Text Solution

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The correct Answer is:
y = 0.1 sin (4t +`pi`/4)
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A point particle of mass 0.1kg is executing SHM of amplitude 0.1m . When the particle passes through the mean position, its kinetic energy is 8xx10^-3J . Obtain the equation of motion of this particle if the initial phase of oscillation is 45^@ .

A point particle of mass 0.1 kg is executing SHM of amplitude 0.1m. When the particle passes through the mean position, its kinetic energy is 8xx10^(-3)J . Obtain the equation of motion of the particle if the initial phase of oscillation is 45^(@)

Knowledge Check

  • A point particle of mass 0.1 kg is executing S.H.M. of amplitude of 0.1 m. When the particle passes through the mean position, its kinetic energy is 8 x× 10^(-3) Joule. Obtain the equation of motion of this particle if this initial phase of oscillation is 45^@ .

    A
    `Y =0.1 sin (pm 4t + pi/4)`
    B
    `Y =0.2 sin (pm 4t + pi/4)`
    C
    `Y =0.1 sin (pm 2t + pi/4)`
    D
    `Y =0.2 sin (pm 2t + pi/4)`

  • A particle of mass 100 g is executing S.H.M. with amplitude of 10 cm. When the particle passes through the mean position at t = 0. Its kinetic energy is 8 mJ. The equation of simple harmonic motion, if initial phase is zero is

    A
    x = 0.1 sin 4t
    B
    x = 0.1 cos 4t
    C
    x = 0.1 sin 2t
    D
    x = 0.1 cos 2t

  • In the equation of motion of a particle y = 0.5 sin (0.3t + 0.1) , the initial phase of motion is -

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    0.1
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    D
    (0.3t + 0.1)

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