When a particle is performing SHM of time period `T_1` the force acting on it is `F_1` for a certain displacement. When the same particle is performing SHM of time period `T_2` the force acting is `F_1` for the same displacement. What will be the time period of the particle when a combined force of `F_1` and `F_2` produces the same displacement in SHM?

Answer

`T =(T_1 T_2)/(sqrt(T_1^2 + T_2^2))`

Explanation

`F_1 = K_1 x , F_2 = - K_2 x, F_1 + F_2 =- kx implies K = K_1 + K_2`
`m omega^2 = m omega_1^2 + m omega_2^2`
`implies (1)/(T_2) = 1/(T_1^2) + 1/(T_2^2) ` (or) `T^2 = (T_1^2 T_2^2)/(T_1^2 + T_2^2) and T= (T_1 T_2)/(sqrt(T_1^2 + T_2^2))`