A particle moves with simple harmonic 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. time period of oscillations is `6tau`
B. amplitude of motion is 3a
C. time period of oscillations is `8tau`
D. amplitude of motion is 4a.

Answer

A

Explanation

`A(1-cos omega tau)=a,A(1-cos 2omega tau)=3a`
`cos omega tau=(1-(a)/(A)),cos 2 omega tau=(1-(3a)/(A))`
`2(1-(a)/(A))^(2)-1=1-(3a)/(A)`
Solving the equation `(a)/(A)=(1)/(2)impliesA=2a`
`cos omega tau=(1)/(2),T=6tau`
So correct choice is (a).