A particle executes SHM such that, the maximum velocity during the oscillation is numerically equal to half the maximum acceleration. What is the time period?

If the maximum acceleration of a particle performing S.H.M. is numerically equal to twice the maximum velocity then the period will be

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

The maximum velocity of a particle executing SHM is 100cms^(-1) and the maximum acceleration is 157cms^(-2) determine the periodic time

A particle executing linear SHM has a maximum velocity of 40 cm ^(-1) and a maximum acceleration of 50 cms^(-2) . Find its amplitude and the period of oscillation.

A particle has maximum velocity 100 cm/sec. and maximum acceleration 157 cm//sec^2 and particle is an SHM. What is the time period.

A particle is executing linear S.H.M. with a maximum velocity of 50 cm/s and a maximum acceleration of 40 cm//s^(2) . Calculate the amplitude and the period of oscillation.

A particle executing linear SHM has a maximum velocity of 40 cm s^(-1) and a maximum acceleration of 50 cm s^(-2) . Find is amplitude and the period of oscillaton. Maximum velocity,

A particle executes SHM along a straight line 4cm long. When the displacement is 1cm its velocity and acceleration are numerically equal. The time period of SHM is