A particle of mass 0.1 kg is executing simple harmonic motion of amplitude 0.1 m. When the particle passes through the mean position, its kinetic energy is `8xx 10^(-3) J`.
If the initial phase is `45^@` , the equation of its motion is (Assume, x(t) as the position of the particle at time t . )

Options

A. `x(t) = 0.1 sin (4t+pi/4)`
B. `x(t) = 0.1 sin (16t +pi/4)`
C. `x(t) =0.1 sin (2(t+pi/4))`
D. `x(t) = 0.1 sin (2t +pi/4)`

Answer

A

Explanation

Key Idea general equation of SHM is given as, `x(t) =A sin (omega t +phi)`
Given , mass of particle m = 0.1 kg aplitude of SHM , A =0.1 m initial phase - difference,
` phi = 45^@ or pi/ 4` and kinetic energy at the mean position
`KE _(max) = 8 xx 10^(-3) J`
`because` Kinetic energy at mean position is maximum. `(because x =0)`
So, `KE _(max) =1/2 m omega ^2 A^2 = 8xx 10^(-3) `
`rArr 1/2 xx 0.1 xx omega^2 xx (0.1) ^2 = 8xx 10^(-3) rArr omega = sqrt(16) = 4 `
Now, the equation of SHM is
`x(t) = A sin (omega t + phi) = 0.1 sin (4t+pi/4)`