The maximum acceleration as time period of motion of a particle executing SHM is `5.0m//s^2` and 6.28s respectively. Find its
acceleration when velocity is `4m//s`.

The time period and amplitude of a particle executing SHM are 10 s and 0.12 m respectively. Find its velocity at a distance 0.04 m from its position of equilibrium.

If the displacement and the acceleration of a particle executing SHM at any instant are x and a respectively, then the time period of that motion will be

Time period and amplitude of a particle executing SHM are 2 s and 5 cm respectively. Determine its velocity and acceleration at a distacne of 3 cm from the equilibrium position.

If the mass of a particle executing SHM is m and its angular frequency is omega , then time period of its oscillation will be

The equation of motion of a particle executing SHM is x=asin(omegat+pi/6) with time period T. Find the time interval at which the velocity is being half of its maximum value.

The maximum acceleration of a particle executing simple harmonic motion is 0.296m*s^(-2) , its time period is 2s and the displacement from the equilibrium position at the beginning of its motion is 0.015m. Determine the equation of motion of the particle.