A particle executing SHM possesses velocities `v_(1)andv_(2)` when it is at distances `x_(1)andx_(2)` respectively from its mean position. Show that, the time period of oscillation is given by `T=2pi((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))^(1//2)`.

Explanation

We know, `v=omegasqrt(A^(2)-x^(2))`.
According to the question,
`v_(1)=omegasqrt(A^(2)-x_(1)^(2))or,v_(1)^(2)=omega^(2)(A^(2)-x_(1)^(2))" "...(1)`
Also, `v_(2)=omegasqrt(A^(2)-x_(2)^(2))or,v_(2)^(2)=omega^(2)(A^(2)-x_(2)^(2))" "...(2)`
Subtracting (2) from (1) we get,
`v_(1)^(2)-v_(2)^(2)=omega^(2)(x_(2)^(2)-x_(1)^(2))`
or, `omega^(2)=(v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2))or,omega=((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))^(1/2)`
We know, `T=(2pi)/omega=(2pi)/(((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))^(1/2))=2pi((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))^(1/2)`