Question

A particle is moving in SHM in a straight line. When the distance of the particle from equilibrium position has values `x_(1)` and `x_(2)` , the corresponding values of velocities are `u_(1)` and `u_(2)`. Show that time period of vibration is `T=2pi[(x_(2)^(2)-x_(1)^(2))/(u_(1)^(2)-u_(2)^(2))]^(1//2)`

Answer 1

As, `V^(2)=omega^(2)(a^(2)-y^(2)),` so
`u_(1)^(2)=omega^(2)a^(2)-omega^(2)x_(1)^(2)` …(i)
And `u_(2)^(2)=omega^(2)a^(2)-omega^(2)x_(2)^(2)` …(ii)
Subtracting (ii) from (i), we get,
`u_(1)^(2)-u_(2)^(2)=omega^(2)(x_(2)^(2)-x_(1)^(2))`
or `u_(1)^(2)-u_(2)^(2)=(4pi^(2))/(T^(2))(x_(2)^(2)-x_(1)^2)`
or `T=2pi[(x_(2)^(2)-x_(1)^(2))/(u_(1)^(2)-u_(2)^(2))]^(1//2)`