The displacement of a particle from its mean position (in metre) is given by
`y=0.2 "sin" (10 pi t+ 1.5 pi) "cos" (10 pi t+1.5 pi)`
The motion of the particle is

Options

A. periodic but not simple harmonic motion
B. non-periodic
C. simple harmonic motion with period of 0.1 s
D. simple harmonic motion with period of 0.2 s

Answer

C

Explanation

`y=0.2 "sin" (10 pi t+1.5 pi) "cos" (10 pi t +1.5 pi)`
U"sin"g `"sin" 2 theta=2 si theta "cos" theta`, we can write this equation as
`y=0.2xx(1)/(2) "sin" 2 [ 10 pi t +1.5 pi]=0.1xx "sin" (20 pi t+ 3pi)`
Comparing this with `y=A "sin" (omega t+ alpha)` we get
`omega =20 pi =(2pi)/(T)`
`therefore T=(2pi)/(omega)=(2pi)/(20 pi)=(1)/(10)=0.1s`
Thus the equation represents a S.H.M. of period 0.1 s.