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The displacement of a particle executing...

The displacement of a particle executing `SHM` at any time `t` (Seconds) is `x =0.01sin 100pi (t = 0.005)` then its time period will be

A

0.2 s

B

0.1 s

C

0.06 s

D

0.02 s

Text Solution

Verified by Experts

The correct Answer is:
D
By comparing with general equation
`x = A sin (omega t + phi)`
we get `omega = 100 pi`
`(2pi)/(T) = 100 pi rArr T = 0.02 s`
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Knowledge Check

  • The displacement of a particle executing S.H.M is given by x= 0.01 sin 100pi (t + 0.05) . The time period is

    A
    0.01s
    B
    0.02s
    C
    0.1 s
    D
    0.2 s

  • The displacement of a particle executing S.H.M. is given by x = 0.34 sin (300 t + 0.68)m. Then its frequency is

    A
    `(300)/(pi)Hz`
    B
    `(300)/(2pi)Hz`
    C
    `(150)/(2pi)Hz`
    D
    `300 Hz`

  • If the displacement of a particle executing S.H.M. is given by x = 0.24 sin (400 t + 0.5)m, then the maximum velocity of the particle is

    A
    `24 m//s`
    B
    `48 m//s`
    C
    `96 m//s`
    D
    `72 m//s`
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