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 oscillation is `6tau`
C. Amplitude of motion is `3a`
D. Time period of oscillation is `8tau`

Answer

B

Explanation

`x = A sin (omegat + pi//2)`
`x = A cos omegat`
`A - a = A cos omegat`
`cos omegat = (A - a)/(A), cos 2 omegat = (A - 3a)/(A)`
`2 cos^(2)omegatau - 1 = (A - 3a)/(A) = 2((A-a)/(A))^(2) - 1`
`2(A - a)^(2)-A^(2) = (A) (A - 3a)`
`2A^(2) + 2a^(2) - 4Aa - A^(2) = A^(2) - 3aA`
`2a^(2) - 4Aa = - 3aA`
`2a^(2) = Aa`
`A = 2a`
`cos omegatau = (a)/(2a) = (1)/(2) , omegatau = (pi)/(3)`
`(2pi)/(omega) = T, T = (2pi.3pi)/(pi) , T = 6tau`