The motion of a particle executing simple harmonic motion is given by X= 0.01 sin 100π (t + 0.05), where X is in metres and t in second. The timeperiod in second is

A particle of mass 0.1 kg0.1 kg is executing simple harmonic motion of amplitude 0.1 m0.1 m. When the particle passes through the mean position, its kinetic energy is 8×10−3 J8×10-3 J. If the initial phase is 45°,45°, the equation of its motion is (Assume, x (t)𝑥 𝑡 as the position of the particle at time t𝑡)

A particle undergoes simple harmonic motion of amplitude x0 and frequency f . What is theaverage speed of the particle during one oscillation?

The function x = (6.0 m) cos [(3π rad/s)t+π/3 rad] gives the simple harmonic motion of a body. The speed as it passes through the equilibrium position is _____ m/s. Round your answer to the nearest tenth.

A particle executes simple harmonic motion with an amplitude of 4 cm4 cm. At the mean position, the velocity of the particle is 10 cm s−110 cm s-1. The distance of the particle from the mean position when its speed becomes 5 cm s−15 cm s-1 is

The function x = (6.0 m) cos [(3π rad/s)t+π/3 rad] gives the simple harmonic motion of a body. The magnitude of the acceleration at an endpoint is _____ m/s². Round to the nearest whole number.