particle of mass 1 kg1 kg is hanging from a spring of force constant 100 N m−1.100 N m-1. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T.𝑇. The minimum time when the kinetic energy and potential energy of the system will become equal, is Tn.𝑇𝑛. The value of n𝑛 is _______

A body of mass 1 kg is executing simple harmonic motion. Its displacement 𝑦 cm at 𝑡 seconds is given by, 𝑦=6sin100𝑡+𝜋4. Its maximum kinetic energy is

A particle executes simple harmonic oscillation with an amplitude A𝐴. The period of oscillation is T𝑇. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position 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)

The motion of a block-spring system is described by x = A sin (ωt). Find ω knowing that the potential energy equals the kinetic energy for the first time at t = 2.00 s.

A 1.00 x 10-2 kg block is resting on a horizontal frictionless surface and is attached to ahorizontal spring whose spring constant is 124 N/m. The block is shoved parallel to the springaxis and is given an initial speed of 8.00 m/s, while the spring is initially unstrained. What isthe amplitude of the resulting simple harmonic motion?