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A particle stars SHM at time t=0. Its am...

A particle stars SHM at time t=0. Its amplitude is A and angular frequency is `omega.` At time t=0 its kinetic energy is `E/4`. Assuming potential energy to be zero and the particle can be written as (E=total mechanical energy of oscillation).

A

x=A sin `omega t +pi//6`

B

x=A sin `(omega t +5pi/6)`

C

x=A sin `(omega t +7pi/6)`

D

x=A sin `(omega t +4pi/6)`

Text Solution

Verified by Experts

The correct Answer is:
D
`(1)/(2)momega^(2)x^(2)=(E)/(4)=(1)/(4)xx(1)/(2)momega^(2)A^(2)`
`Rightarrow x=_(-)^(+)(A)/(2) Rightarrow v=_(-)^(+)(omegaA)/(2)Rightarrowx=_(-)^(+)(sqrt3A))/(2)`
`phi= (pi)/(3),(2pi)/(3),(4pi)/(3),(5pi)/(3) Rightarrow x=(A)/(2)sin(omegat+phi)` where `phi= (pi)/(3),(2pi)/(3),(4pi)/(3),(5pi)/(3)`
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  • A particle of mass m executing SHM with amplitude A and angular frequency omega . The average value of the kinetic energy and potential energy over a period is

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