A mass m is suspended from the two coupled springs connected in series. The force constant for springs are `k_(1) "and" k_(2).` The time period of the suspended mass will be

Options

A. `T=2pisqrt(((m)/(K_(1)+K_(2))))`
B. `T=2pisqrt(((2m)/(K_(1)+K_(2))))`
C. `T=2pisqrt(((m(K_(1)+K_(2)))/(K_(1)K_(2))))`
D. `T=2pisqrt(((mK_(1)K_(2))/(K_(1)+K_(2))))`

Answer

C

Explanation

`k_(eq)=(K_(1)K_(2))/(K_(1)+K_(2))` [as spring connected in series]
`T=2pisqrt((m(K_(1)+K_(2)))/(K_(1)K_(2)))`.