Question

A particle executing SHM has velocities v₁ and v₂ when it is at distance x₁ and x₂ from the centre of the path. Show that the time period is given by `T=2pisqrt((x_2^2-x_1^2)/(v_1^2-v_2^2))`

Answer 1

For a particle in SHM velocity at any instant is `v =sqrt(a^2-x^2)`

`v_1 =omegasqrt(a^2-x_1^2)and v_2= sqrt(a^2-x_2^2)`

`v_1^2=omega^2(a^2-x_1^2) and v_2^2=omega^2(a^2-x_a^2)`

`v_1^2/omega^2=(a^2-x_1^2)       ...(1)`

`v_2^2/omega^2 =(a^2-x_2^2)     ...(2)`

Subtracting equation (1) from equation (2) gives,

`v_1^2/omega^2-v_2^2/omega^2=x_2^2-x_1^2`

`(v_1^2-v_2^2)/omega^2=x_2^2-x_1^2`

`omega^2 = (v_1^2 - v_2^2)/(x_2^2-x_1^2)`

`omega=sqrt((v_1^2 - v_2^2)/(x_2^2-x_1^2))`

But `omega = (2pi)/T`

`(2pi)/T=sqrt((v_1^2 - v_2^2)/(x_2^2-x_1^2))`

`T = 2pisqrt((x_2^2-x_1^2)/(v_1^2-v_2^2))`