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A particle performing SHM starts equi...

A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is `pi m//s`. Amplitude of oscillation is
`(cos 45^(@) = 1/(sqrt(2)))`

A

`2sqrt2`m

B

`4sqrt2`m

C

`6sqrt2`m

D

`8sqrt2`m

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The correct Answer is:
D
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