For a particle in `SHM` the amplitude and maximum velocity are `A` and `V` respectively. Then its maximum acceleration is

At particle is executing SHM with amplitude A and has maximum velocity v_(0) . Find its speed when it is located at distance of (A)/(2) from mean position.

At particle is executing SHM with amplitude A and has maximum velocity v_(0) . Find its speed when it is located at distance of (A)/(2) from mean position.

If a particle executing S.H.M. with amplitude A and maximum velocity vo, then its speed at displacement A/2 is

A particle is performing simple harmonic motion with amplitude A and angular velocity omega . The ratio of maximum velocity to maximum acceleration is

For a particle in S.H.M. if the maximum acceleration is a and maximum velocity is v then amplitude is

The displacement of a particle performing SHM x= 3sin 2t+4 cos 2t . The amplitude and maximum velocity of a particle respectively are………….

A particle is executing S.H.M. with amplitude 'A' and maximum velocity V_(m) . The displacement at which its velocity is half of the maximum velocity is

A particle is executing SHM with an amplitude of 2 m. The difference in the magnitudes of its maximum acceleration and maximum velocity is 4. The time period of its oscillation is

A particle executes a S.H.M. of amplitude A and maximum velocity V_(m) then its speed at displacement A//2 is