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

A particle executing SHM with time period T and amplitude A. The mean velocity of the particle averaged over quarter oscillation, is

A particle executing SHM with time period T and amplitude A. The mean velocity of the particle averaged over quarter oscillation, is

A particle executes SHM with a time period of 2 s and amplitude 5 cm. What will be the displacement and velocity of that particle at t = 1/3 ?

A particle executes SHM with a time period T . The time period with which its potential energy changes is

Amplitude of a particle executing SHM is a and its time period is T. Its maximum speed is

Amplitude of a particle executing SHM is a and its time period is T. Its maximum speed is

A particular executes SHM with a time period of 16s. At time t=2s , the particle crosses the mean position while at t= 4 , its velocity is 4ms^(-1) Find its a amplitude of motion.

A SHO has amplitude A and time period T . The maximum velocity will be

A particle is executing SHM with time period T. If time period of its total mechanical energy isT' then (T')/(T) is