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The velocity of a particle describing SH...

The velocity of a particle describing SHM is `16 cm s^(-1)` at a distance of 8 cm from mean position and `8 cm s^(-1)` at a distance of 12 cm from mean postion. Calculate the amplitude of the motion.

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A particle is executing simple harmonic motion. The velocity of the particle is 10 m s^(-1) . What at a distance of 5 cm from mean position and 8 cm s^(-1) when at a distance of 12 cm from mean position. What will be the amplitude of motion ?

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Knowledge Check

  • The acceleration of a particle performing S.H.M. is 12 cm // sec^(2) cm at a distance of 3 cm form the mean position. Its period is

    A
    0.5 sec
    B
    1.0 sec
    C
    2.0 sec
    D
    3.14 sec

  • The phase of a particle performing S.H.M. when the particle is at a distance of amplitude from mean position is

    A
    `pi//2`
    B
    `pi`
    C
    `3pi//2`
    D
    odd multiple of `pi//2`

  • The acceleration of a particle executing SHM at a distance of 3 cm from equilibrium position is 5 cm//s^(2) . Its acceleration at a distance of 2 cm from equilibrium position is

    A
    `10//3 cm//s^(2)`
    B
    `10 cm//s^(2)`
    C
    `7.5 cm//s^(2)`
    D
    `4.5 cm//s^(2)`

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