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A particle starts performing simple harm...

A particle starts performing simple harmonic motion. Its amplitude is `A`. At one time its speed is half that of the maximum speed. At this moment the displacement is

A

`(sqrt(2)A)/(3)`

B

`(sqrt(3)A)/(2)`

C

`(2 A)/sqrt(3)`

D

`(3A)/sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`v = omega sqrt (A^(2) - x^(2))`
`:. (omega A)/(2) = omega A sqrt (A^(2) - x^(2))` or `x = (sqrt3)/(2)A`
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