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

Options

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

Answer

B

Explanation

Given, `y=3cos((pi)/(4)-2omegat)` …(i)
velocity , `V=(dy)/(dt)=3xx2omega sin((pi)/(4)-2omegat)`
Acceleration, `A=(dV)/(dt)=-4omega^(2)xx3cos((pi)/(4)-2omegat)=-4omega^(2)y`
As `Aporp y` and `-ve` sign shows that it is directed towards equilibrium (or mean position), hence particle will execute SHM. Compare (i) with equation
`y=r cos(phi-omega^(')t)`
We have ` omega=2omega` or `(2pi)/(T^('))=2omega` or ` T^(')=(pi)/(omega)`