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A particle of mass 0.1 kg is executing S...

A particle of mass 0.1 kg is executing SHM of amplitude 0.1 m . When the particle passes through the mean position , its KE is `8 xx 10^(-3)` J. Find the equation of motion of the particle if the initial phase of oscillation is `45^(@)`

A

`y = 0.1 cos (3t + pi//4)`

B

`y = 0.1 sin (6t + pi//4)`

C

`y = 0.1 sin (4t + pi//4)`

D

`y = 0.1 cos (4t + pi//4)`

Text Solution

Verified by Experts

The correct Answer is:
C
At mean position `KE = 1/2momega^(2)a^(2) = 8 xx 10^(-3) rArr omega^(2) = 4"rad"//"sec"`
Now at equation of SHM be `y = 0.1sin(4t+phi)`, At `t = 0, phi = 45^(@) = (pi)/(4)`. Therefore `y = 0.1sin(4t + (pi)/(4))`
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