A particle executing SHM along a straight line has zero velocity at point A and B whose distance from 0 on the same line OAB are a and b respectively if the velocity at the mid point between A and B 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

According to the question

Amplitude of particles executing
Simple harmonic motion (SHM) along a straigh line AB is `(a ) = (b-a)/(2)`
Velocity of particle v= Amplitude xx Oscillation
`:. , v=a epsilon `
`v= ((b-a)/(2)) epsilon " or " epsilon = (2 v)/(b-a)`
Putting the value of `epsilon` from Eq (i ) to above formula
`rArr T = (2pi)/( 2 v) xx (b-a) rArr T = (b-a)/(v ) pi`
So the time period of a particle executing SHM along a straight line from points A to B is
`T= (b-a)/(V ) pi`