A particle executes SHM in a straight line. In the first second starting from rest it travels 'a' distance a and in the next second a distance 'b' in the same direction . The amplitude of S.H.M will be

Options

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

Answer

A

Explanation

`x=A cos wt (` as it starts from rest at `t=0)`

`A-a=A cos w ……(1)`
`A-(a+b)=A cos 2w ….(2)`
Solving `(1)` and `(2)` for `A` we get `A =(2a^(2))/(3a-b)`