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A particle executes simple harmonic moti...

A particle executes simple harmonic motion with an amplitude A. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude form the equilibrium position is………..

A

`(T)/(2)`

B

`(T)/(4)`

C

`(T)/(8)`

D

`(T)/(12)`

Text Solution

Verified by Experts

The correct Answer is:
D
Displacement at mean position
`y= a sin ((2pi)/(T))t`
For `y= (a)/(2), (a)/(2)= a sin ((2pi)/(T))t`
`therefore (1)/(2)= sin (pi)/(6) = sin [((2pi)/(T))t]`
`therefore t((2pi)/(T))= (pi)/(6)`
`therefore t= (T)/(12)` is minimum.
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Knowledge Check

  • A particle executes simple harmonic motion with frequency f. The frequency of its potential and kinetic energy is………

    A
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    B
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