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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

`2sqrt2m`

B

`4sqrt2m`

C

`6sqrt2m`

D

`8sqrt2m`

Text Solution

Verified by Experts

The correct Answer is:
D
Displacement of the particle `=x=A sin omegat`
Velocity of the particle `=v=(dx)/(dt)=A omega cos omegat`
`v=pi m//s, T=16s, omega=(2pi)/(T)=(2pi)/(16)"rad/s"`
`therefore pi=A xx (pi)/(8)xx cos""(pi)/(8)xx2`
`therefore 1=(A)/(8)cos ""pi/4=(A)/(8).(1)/(sqrt2)`
`therefore A=8sqrt2m`
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