A mass is suspended separately by two springs of spring constants `k_(1)` and `k_(2)` in successive order. The time periods 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 time period of oscillations is T. The correct relations 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

2

Explanation

`T = 2pisqrt((M)/(K))`
`K= (4 pi ^(2)M)/(T^(2))`
`K_(1) = (4 pi^(2)M)/(T_(1)^(2)),K= (4 pi^(2)M)/(T_(2)^(2))`
Spring are in parallel combination
`K_(eq)= K_(1) + K_(2)`
`(1)/(T_(eq)^(2))=(1)/(T_(1)^(2))+ (1)/(T_(2)^(2))`
`T_(eq)^(-2) = T_(1)^(-2) + T_(2)^(-2)`