The displacement of a particle is represented by the equation `y=3cos((pi)/(4)-2omegat).` The motion of the particle is
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
Text Solution
Verified by Experts
The correct Answer is:
B
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)`.
