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_(1)`
B. `T_(2)`
C. `sqrt(T_(1)^(2)+T_(2)^(2))`
D. `(T_(1)T_(2))/(sqrt(T_(1)^(2)+T_(2)^(2)))`

Answer

D

Explanation

`T_(1)^(2)=(4pi^(2)m)/(k_(1))andT_(2)^(2)=(4pim)/(k_(2))`
when springs are connected in parallel effective force constant
`k_(p)=k_(1)+k_(2)`
`therefore (1)/(T_(P)^(2))=(1)/(T_(1)^(2))+(1)/(T_(2)^(2))=(T_(2)^(2)+T_(1)^(2))/(T_(1)^(2)T_(2)^(2))`