A particle moves with simple harmomonic motion in a straight line. In first `tau s`, after starting from rest it travels a distance a, and in next `tau s` it travels 2a, in same direction, then :-

Options

A. amplitude of motion is 4a
B. time period of oscillations is 6`pi`
C. time period of oscillations is 6`pi`
D. time period of oscillations

Answer

B

Explanation

As the particle starts from rest so we choose
`x(t)=A cos omegat`
At `t=0,x=A`
When `t=tau, x=A-a`
when `t=2tau,x=A-3a thereforeA-a=Acos omegat`
and `A-3a=Acos 2omega tau= A(2cos^(2)omegat-1)`
`implies(A-3a)=A[2((A-a)/A)^(2)-1]implies (A-3a)/(A)=2((A-a)/A)^(2)-1`
on solving ,A=2a
Now ` A-a =Acos omegatau,` `impliescos omegatau =-1/2="cos" (pi)/3 impliesomegatau=pi/3 implies`