A mass `m` is suspended separately by two different springs of spring constant `k_(1)` and `k_(2)` given the time period `t_(1)` and `t_(2)` respectively. If the same mass `m` is shown in the figure then time period `t` is given by the relation

Options

A. `t = t_(1) + t_(2)`
B. `t =( t_(1)t_(2))/(t_(1) + t_(2))`
C. `t^(2) = t_(1)^(2) + t_(2)^(2)`
D. `t^(-2) = t_(1)^(-2) + t_(2)^(-2)`

Answer

D

Explanation

`t_(1) = 2pi sqrt((m)/(k_(1)))` and `t_(2) = 2pi sqrt((m)/(k_(2)))`
Equivalant constant for shown combination is , `(K_(1) +K_(2)` so time period `t` is given by
`t= 2pi sqrt((m)/(K_(1) + K_(2)))`
By solving these equations we get `T^(2) = t_(1)^(2) + t_(2)^(2)`