(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)`

Options

A. `2pi sqrt((x_(1)^(2) + x_(2)^(2))/(V_(1)^(2) + V_(2)^(2)))`
B. `2pi sqrt((x_(2)^(2) - x_(1)^(2))/(V_(1)^(2) - V_(2)^(2)))`
C. `2pi sqrt((V_(1)^(2) + V_(2)^(2))/(x_(1)^(2) + x_(2)^(2)))`
D. `2pi sqrt((V_(1)^(2) - V_(2)^(2))/(x_(1)^(2) - x_(2)^(2)))`

Answer

B

Explanation

As we know for particle undergoing `SHM`
`V = omega sqrt(A^(2) + X^(2))`
`v_(1)^(2) = omega^(2)(A^(2) + x_(1)^(2)) and V_(2)^(2) = omega^(2)(A^(2) + x_(2)^(2))`
On subracting the relations
`V_(1)^(2) - V_(2)^(2) = omega^(2)(x_(2)^(2) - x_(1)^(2))`
`omega = sqrt((V_(1)^(2) - V_(2)^(2))/(x_(2)^(2) - x_(1)^(2)))`
`T = 2pi sqrt((x_(2)^(2) - x_(1)^(2))/(V_(1)^(2) - V_(1)^(2)))`