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A pendulum is hung from the rod of a suf...

A pendulum is hung from the rod of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is `20 m"/"s^(2)` at a distance of 5 m from the mean position. The period of oscillation is...........

A

`1s`

B

`2pi s`

C

`2s`

D

`pi s`

Text Solution

Verified by Experts

The correct Answer is:
D
`|a| = W^(2) x`
`20= W^(2) xx 5`
`therefore W^(2) =(20)/(5)= 4`
`therefore W= 2`
`therefore (2pi)/(T)= 2`
`therefore T= (2pi)/(2)`
`therefore T= pi s`.
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KUMAR PRAKASHAN-OSCILLATIONS-SECTION-E (MCQs ASKED IN GUJARAT BOARD AND COMPETITIVE EXAMS)
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  10. When the kinetic energy of the body executing SHM is (1)/(3) of potent...

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  12. When two displacements represented by y(1)= a sin(omega t)" and "y(2)=...

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  14. A particle is executing a simple harmonic motion. Its maximum accelera...

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