When a particle executing SHM is at a distance of 0.02 m from its mean position, then its kinetic energy is thrice its potential energy. Calculate the amplitude of motion of the particle.

Explanation

According to the question, K = 3 U
i.e., `1/2momega^(2)(A^(2)-x^(2))=3xx1/2momega^(2)x^(2)or,A^(2)-x^(2)=3x^(2)`
or, `A^(2)=4x^(2)or,A=|pm2x|=0.04` [since x = 0.02m]
`therefore` Amplitude = 0.04 m.