Two simple harmonic motions with the same frequency act on a particle at right angles ie, along X and Y-axis.  If the two amplitudes are equal and the phase difference is , the resultant motion will be:

A particle moving along the X-axis executes simple harmonic motion; then the force acting on it is given by:

A point particle of mass 0.1 kg is executing S.H.M of amplitude 0.1 m.  When the particle passes through the mean position, its K.E is 8 x 10^-3 J.  The equation of motion of this particle if its initial phase of oscillations is 45^o is:

A particle executes simple harmonic motion of amplitude A.  At what distance from the mean position its kinetic energy is equal to its potential energy ?

The angular velocity and the amplitude of a simple pendulum is and respectively.  At a displacement x from the mean position, if its kinetic energy is T and potential energy is U, then the ration of T to U is:

A simple pendulum is executing simple harmonic motion with a time period T.  If the length of the pendulum is increased by 21%, the increase in the time period is:

The displacement x (in metres) of a particle performing S.H.M is related to time t (in second) as :

A simple harmonic oscillator has an amplitude and time period T.  The time required by it to travel from   x = to x = is:

In a simple harmonic motion, when the displacement is one-half the amplitude, what fraction of the total energy is kinetic ?

Two simple harmonic motions given by,
act on a particle simultaneously, then the motion of particle will be