The amplitude of a particle of a mass 1 kg completing S.H.M. is 0.1 m. If- the kinetic energy of that -body while passing the equilibrium position be` 8 xx 10^-3 J` then find out its time period.

Amplitude of a particle of mass 0.1 kg executing SHM is 0.1 m. At the mean position its kinetic energy is 8xx10^(-3)J . Find the time period of its vibration.

The amplitude of a particle executing SHM is 0.1 m. The acceleration of the particle at a distance of 0.03 m from the equilibrium position is 0.12m*s^(-2) . What will be its velocity at a distance of 0.08 m from the position of equilibrium?

The time period of a particle executing S.H.M. is T sec. Find the time taken by the particle to reach maximum displacement in one side from equilibrium position.

A particle of mass m at rest in acted upon by a force P for a time t. its kinetic energy after an interval t is

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.

A particle of mass m executes SHM with amplitude a and frequency n. What is the average kinetic energy of the particle during its motion from the position of equilibrium to the end?