A point particle if mass `0.1 kg` is executing SHM of amplitude `0.1 m`. When the particle passes through the mean position, its kinetic energy is `8 xx 10^(-3)J`. Write down the equation of motion of this particle when 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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