A particle of mass 2 kg executes simple harmonic motion. The total distance between its two extreme positions is 4 m, and it takes 1 s to move from the mean position to an extreme position. What is the kinetic energy of the particle at the moment its acceleration has a magnitude of ^2 4 m/s ^2 ?

Step-by-step solution

The distance between the two extreme positions is 2A . 2A = 4 m A = 2 m. The time taken to move from the mean position to an extreme position is one-fourth of the time period. T 4 = 1 s T = 4 s. The angular frequency is = 2 T = 2 rad/s. The magnitude of acceleration is given by |a| = ^2 x . Substitute the given acceleration to find the position x : ^2 4 = ( 2 )^2 x ^2 4 = ^2 4 x x = 1 m. The kinetic energy of a particle in SHM is K = 1 2 m ^2 (A^2 - x^2) . K = 1 2 (2) ( 2 )^2 (2^2 - 1^2) K = 1 ^2 4 (4 - 1) = 3 ^2 4