A particle is executing S.H.M. from mean position at 5 cm distance, acceleration is `20 cm//sec^(2)` then value of angular velocity will be

A particle executes SHM in a straight line path .The amplitude of oscillation is 2cm . When the displacement of the particle from the mean position is 1cm the numerical value of acceleration is equal to the numerical value of velocity. Then find the frequency of SHM .

A particle executing a SHM has maximum acceleration at a distance of 0.5 cm from its mean position is 2cm//s^(2) . What will be its velocity when it is at a distance of 1 cm from its mean position.

A particle is executing S.H.M. along 4 cm long line with time period (2pi)/(sqrt2) sec. If the numerical value of its velocity and acceleration is same, then displacement will be

If particle executing S.H.M. has an amplitude of 6 cm, its acceleration at a distance 2 cm from mean position is 8 cm/s^2 , then the maximum speed of the particle will be

The particle is executing S.H.M. on a line 6 cms long. If its velocity at mean position is 12 cm/sec, its frequency in Hertz will be:

The maximum displacement of a particle executing S.H.M. is 1 cm and the maximum acceleration is (1.57)^(2) cm per sec^(2) . Then the time period is

A particle is executing a linear S.H.M. Its velocity at a distance x from the mean position is given by v^2=144-9x^2 . The maximum velocity of the particle is

A particle is executing a linear S.H.M. Its velocity at a distance x from the mean position is given by v^2=144-9x^2 . The maximum velocity of the particle is