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How long after the beginning of motion is the displacment of a harmonically oscillation particle equal to one half its amplitude if the period is `24s` and perticle starts from rest.

A

`12s`

B

`2s`

C

`4s`

D

`6s`

Text Solution

Verified by Experts

The correct Answer is:
C
`y = a cos omegat`
`(a)/(2) = a cos omegat`
`omegat = (pi)/(3)`
`(2pi)/(24)t = (pi)/(3)`
`t = 4 sec`.
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RESONANCE-SIMPLE HARMONIC MOTION -Exercise- 1, PART - II
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  3. A particle performing SHM takes time equal to T (time period of SHM) i...

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  4. A particle executing linear SHM. Its time period is equal to the small...

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  6. Two SHM's are represented by y = a sin (omegat - kx) and y = b cos (om...

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  7. How long after the beginning of motion is the displacment of a harmoni...

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  8. The magnitude of average accleration in half time period in a simple h...

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  10. Two particles execute SHMs of the same amplitude and frequency along t...

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  11. A mass M is performing linear simple harmonic motion. Then correct gra...

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  13. The KE and PE , at is a particle executing SHM with amplitude A will b...

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  16. Acceleration a versus time t graph of a body in SHM is given by a curv...

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