The velocity of a particle in `S.H.M.` at positions `x_(1)` and `x_2` are `v_(1)` and `v_(2)` respectively. Determine value of time period and amplitude.

Explanation

`V= omega sqrt(A^2 - x^2)`
`V^2= omega^2 (A^2-X^2)`
At position `x_1` velocity `V_1^2= omega^2 (A^2 - x_1^2)…...(A)`
At position `x_2` velocity `v_2^2= omega^2 (A^2 - x_2^2)….......(B)`
(i) Subtracting (B) from (A)
`V_1^2-V_2^2= omega^2 (x_2^2- x_1^2) rArr omega= sqrt((V_1^2- V_2^2)/(x_2^2- x_1^2))`
Time period `T= (2pi)/(omega) rArr T= 2pi sqrt((X_2^2- x_1^2)/(V_1^2 - V_2^2))`
(ii) Dividing (A) by (B)
`(V_1^2)/(V_2^2) = (A^2-X_1^2)/(A^2- X_2^2) rArr V_1^2 A^2- V_1^2 X_2^2 = V_2^2 A^2 - V_2^2 X_1^2`
So `A^2 (V_1^2- V_2^2)= V_1^2 X_2^2 - V_2 ^2 x_1 ^2`
`rArr A= sqrt((V_1^2 X_2^2 - V_2^2 x_1^2)/(V_1^2- V_2^2))`