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A particle performing SHM is found at it...

A particle performing SHM is found at its equilibrium position at t = 1s and it is found to have a speed ` 0.25 m//s` at t=2s. If the period of oscillation is 6. Calculate amplitude of oscillation.

A

`(3)/(2pi)m`

B

`(3)/(4pi)m`.

C

`(6)/(pi)m`

D

`(3)/(8pi)m`

Text Solution

Verified by Experts

The correct Answer is:
A
`x = a sin (omegat + phi)`
at `t = 1s, x = 0 = a sin(omegat + phi) rArr phi = -omega`
`:' v = a omega cos (omegat + phi)`
`:'` At `t = 2s, 1/4 = aomega cos (2omega + phi)`
`rArr 1/4 = a ((2pi)/(6)) cos (omega) = (api)/(3)cos ((pi)/(3)) = (api)/(3)((1)/(2)) rArr a = (3)/(2pi)`
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ALLEN -SIMPLE HARMONIC MOTION-Exercise-01
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