(a) The motion of the particle in simple harmonic motion is given by `x = a sin omega t`. If its speed is `u`, when the displacement is `x_(1)` and speed is `v`, when the displacement is `x_(2)`, show that the amplitude of the motion is
`A = [(v^(2)x_(1)^(2) - u^(2)x_(2)^(2))/(v^(2) - u^(2))]^(1//2)`
(b) A particle is moving with simple harmonic motion is a straight line. When the distance of the particle from the equilibrium position has the values ` x_(1)` and `x_(2)` the corresponding values of velocity are `u_(1)` and `u_(2)`, show that the period is
`T = 2pi[(x_(2)^(2) - x_(1)^(2))/(u_(1)^(2) - u_(2)^(2))]^(1//2)`

Explanation

`T = 2pi ((x_(2)^(2) - x_(1)^(2))/(v_(1)^(2) - v_(2)^(2)))^(1//2)`