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A particle is executing simple harmonic ...

A particle is executing simple harmonic motion with a period of T seconds and amplitude a metre . The shortest time it takes to reach a point `a/sqrt2` from its mean position in seconds is

A

T

B

`(T)/(4)`

C

`(T)/(8)`

D

`(T)/(16)`

Text Solution

Verified by Experts

The correct Answer is:
C
(c)`y=a "sin"(2pi)/(T)t`
`implies(A)/(sqrt(2))=a "sin"(2pi)/(T).t`
`implies"sin" (2pi)/(T)t=(1)/(sqrt(2))="sin"(pi)/(4)`
`implies (2pi)/(T)t=(pi)/(4)`
`therefore t=(T)/(8)`
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