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The amplitude and time period of a parti...

The amplitude and time period of a particle of mass 0.1 kg executing simple harmonic motion are 1m and 6.28s, respectively. Find its (i) angular frequency, (ii) acceleration and (iii) velocity at a displacement of 0.5m.

A

`1 "rad"//"sec",-1ms^(-2), sqrt(3)ms^(-1)`

B

`0.5 "rad"//"sec",-0.5ms^(-2), (sqrt(3))/(2) ms^(-1)`

C

`1 "rad"//"sec",-0.5 ms^(-2), (sqrt(3))/(2) ms^(-1)`

D

`1 "rad"//"sec",-0.5 ms^(-2), sqrt(3)ms^(-1)`

Text Solution

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The correct Answer is:
C
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Knowledge Check

  • The amplitude and time period of a particle of mass 0.1 kg executing simple harmonic motioni are 1 m and 6.28 s, respectively. Find its (i) angular frequency (ii) acceleration and (iii) velocity at a displacement of 0.5 m

    A
    1 rad/sec,`-1ms^(-2),sqrt(3)ms^(-1)`
    B
    0.5 rad/sec , `-0.5ms^(-2),(sqrt(3))/2ms^(-1)`
    C
    1 rad/sec , `-0.5ms^(-2),(sqrt(3))/2ms^(-1)`
    D
    1 rad/sec , `-0.5ms^(-2),sqrt(3)ms^(-1)`

  • The amplitude and time period of a particle of mass 0.1 kg executing simple harmonic motioni are 1 m and 6.28 s, respectively. Find its (i) angular frequency (ii) acceleration and (iii) velocity at a displacement of 0.5 m

    A
    1 rad/sec,`-1ms^(-2),sqrt(3)ms^(-1)`
    B
    0.5 rad/sec , `-0.5ms^(-2),(sqrt(3))/2ms^(-1)`
    C
    1 rad/sec , `-0.5ms^(-2),(sqrt(3))/2ms^(-1)`
    D
    1 rad/sec , `-0.5ms^(-2),sqrt(3)ms^(-1)`

  • At some instant, velocity and acceleration of a particle of mass 'm' in SHM are 'a' and 'b' respectively. If its angular frequency is omega , then its maximum K.E is

    A
    `1/2 m [a^2 + (b^2)/(omega)]`
    B
    `1/2 m[a^2 - (b^2)/(omega)]`
    C
    `1/2 m[a^2 + (b^2)/(omega^2)]`
    D
    ` m [a^2 + (b^2)/(omega^2)]`

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