A particle executes SHM with a period of T second and amplitude A metre. The particle is at its mean position initially. The shortest time it takes to reach point A√2 metre from its mean position in seconds is
A
T
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B
T4
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C
T8
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D
T16
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Solution
The correct option is CT8 Let y=A sin ωt
i.e. A√2=A sin ωt ⇒sin ωt=1√2=sin π4
i.e. ωt=π4
But ω=2πT
Therefore, t=T8