(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)`

Answer

A

Explanation

(a) `u = omega sqrt (A^(2) - x_(1)^(2))` …(i)
`v = omega sqrt (A^(2) - x_(2)^(2))` …(ii)
Solving Eqs. (i) and (ii), we get the result,
`A or a = sqrt ((v^(2)x_(1)^(2) - u^(2)x_(2)^(2))/(v^(2) - u^(2))`
(b) `u_(1) = omega sqrt (A^(2) - x_(1)^(2))` ...(iii)
` u_(2) = omega sqrt (A^(2) - x_(2)^(2))` ...(iv)
Solving Eqs. (iii) and (iv), we find
`omega = sqrt ((u_(1)^(2) - u_(2)^(3))/(x_(2)^(2) - x_(1)^(2))) = (2pi)/(T)`
`:. T = 2pi sqrt ((x_(2)^(2) - x_(1)^(2))/(u_(1)^(2) - u_(2)^(2))` .