A particle performs SHM in a straight line. In the first second, starting from rest, it travels a distance a and in the next second it travels a distance b in the same side of mean position. The amplitude of the SHM is

Options

A. `(2a^(2))/(3a-b)`
B. `a-b`
C. `2a-b`
D. `a//b`

Answer

A

Explanation

`x = Acosomegat rArr a = A cosomega` and `a + b = Acos2omega`
`rArr a+ b = A[2cos^(2)omega-1] = A[2.(a^(2))/(A^(2))- 1]`
`rArr (2a^(2))/(A) - A = a+b rArr A^(2) + (a+b)A - 2a^(2) = 0`
`rArr A = (-(a+b)+sqrt(a^(2) + b^(2) + 2ab + 8a^(2)))/(2)`