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.

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.

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

Assume a body performing harmonic oscillation, with the oscillation period of T = 6 s; the initial phase is zero. Calculate how long (in seconds) after the beginning of the motion the body displacement from equilibrium position will be one half of the amplitude.

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 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