The displacement of a particle is represented by the equation `y=3cos((pi)/(4)-2omegat).`
The motion of the particle is

Options

A. a simple harmonic with period `2pi//omega`
B. a simple harmonic with period `pi//omega`
C. periodic but not simple harmonic
D. non-periodic

Answer

B

Explanation

The displacement of the particle
`y = 3cos ((pi)/(4)-2omegat)` velocity of the particle
`v = (dy)/(dt) = (d)/(dt) [3cos ((pi)/(4)-2 omegat)]`
`= 6omega sin ((pi)/(4)-2 omegat)`
Acceleration
`a = (dv)/(dt) = (d)/(dt) [6omega sin ((pi)/(4)-2 omegat)]`
`=- 12 omega^(2) cos ((pi)/(4)-2 omegat)=`
`-4 omega^(2) [ 3cos ((pi)/(4) -2 omegat)]`
Here `a =- 4 omega^(2)y =- (2omega)^(2) y`
It means acceleration, `a alpha -y`, the motions is `SHM`. Hence angular frequency of `S.H.M, omega' = 2 omega`
`omega' = 2omega = (2pi)/(T') rArr T' = (2pi)/(2omega) = (pi)/(omega)`
It means the motion is `SHM` with period `(pi)/(omega)`.
`METHOD 2`: Given the equation of displacement of the particle
`y = 3cos ((pi)/(4)-2 omegat)`
`y = 3cos [-(2omegat-(pi)/(4))]`
We know `cos (-theta) = cos theta`
Hence `y = 3 cos (2 omegat -(pi)/(4))`
Compering with `y =a cos (omegat +phi_(0))`
Hence (i) reprecents simple hamronic motion with angular frequency `2omega`.
Hence its time period, `T = (2pi)/(2omega) = (pi)/(omega)`.