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A particle simple harmonic motion comple...

A particle simple harmonic motion completes `1200` oscillations per minute and passes through the mean position with a velocity `3.14 ms^(-1)`. Determine displacement of the particle from its mean position. Also obtain the displacement equation of the particle if its displacement be zero at the instant `t = 0`.

Text Solution

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The correct Answer is:
B, D
The maximum velocity of the particle at the mean position
`v_(max) = Aomega = A(2pin)`
`rArr A = (v_(max))/(2pin) = (3.14)/(2 xx 3.14 xx 20) = 0.025 m`
If at the instant t = 0, displacement be zero so displacement equation is
`y = Asinomegat = Asin2pint = 0.025 sin (40pit) m`
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Knowledge Check

  • The displacement of a particle represented by the equation y= 3 cos((pi)/(4)-2 omega t) . The motion of the particle is

    A
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    B
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  • Acceleration of a particle a= -bx , where x is the displacement of particle from mean position and b is constant. What is the periodic time of oscillation ?

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    B
    `(2pi)/(sqrt(b))`
    C
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    D
    `2 sqrt(((pi)/(b))`

  • The displacement of a particle is represented by the equation y= sin^(3) omega t . The motion is………..

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    non-periodic
    B
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    C
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