A simple pendulum executing `SHM` in a straight line has zero velocity at 'A' and 'B' whose distances form 'O' in the same line `OAB` are 'a' and 'b'. If the velocity half wat between them is 'v' then its time period is

Options

A. `(pi (b+a))/(v)`
B. `pi((b-a)/(v))`
C. `((b+a)/(2v))`
D. `((b-a)/(2v))`

Answer

B

Explanation

A and B are the extreme position `rArr 2A = b - a`
`rArr A = (b - a)/(2)` is the amplitude
Halfway between them is the mean position `v = A omega`
`T = (2pi)/(omega) = (2 pi A)/(v) = (2pi)/(v) ((b-a)/(2)) - (pi (b-a))/(v)`