A mass is suspended separately by two springs of spring constant `k_(1)` and `k_(2)` in successive order. The time period of oscillations in the two cases are `T_(1)` and `T_(2)` respectively .If the same mass be suspended by connecting the two springs in parallel, (as shown in figure ) then the timer period of oscillations is T. The correct relation is

Options

A. `T^(2)=T_(1)^(2)+T_(2)^(2)`
B. `T^(-2)=T_(1)^(-2)+T_(2)^(-2)`
C. `T^(-1)=T_(1)^(-1)+T_(2)^(-1)`
D. `T=T_(1)+T_(2)`

Answer

B

Explanation

`T_(1)=2pisqrt((m)/(k_(1)))`, `T_(2)2pisqrt((m)/(k_(2)))`, `T=2pisqrt((m)/(k_(1)+k_(2)))` `Rightarrow` `(!)/(T^(2))=(1)/(T_(1)^(2))+(1)/(T_(2)^(2))``Rightarrow` `T^(-2)=T_(1)^(-2)+T_(2)^(-2)`