A particle executes `SHM` with time period `T` and amplitude `A`. The maximum possible average velocity in time `T//4` is-

Options

A. `(2A)/(T)`
B. `(4A)/(T)`
C. `(8A)/(T)`
D. `(4sqrt(2)A)/(T)`

Answer

D

Explanation

Maximum possible average velocity will be around mean position.
`T/4 = (2(A//sqrt(2)))/(T//4) = (4sqrt(2)A)/(T)`
OR
`phi = omegat = (2pi)/(T)(T)/(8)= (pi)/(4)`

`cos phi = (x)/(A) rArr cos'(pi)/(4) = (x)/(A) rArr x = (A)/(sqrt(2))`
Average velocity `= ("total displacement")/("total time")`
`= (2x)/(T//4) = (2A//sqrt(2))/(T//4) = (4sqrt(2)A)/(T)`