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

Answer

A

Explanation

When the springs are connected in series, the effective value of the spring constant `K_("eff")=(K_(1)K_(2))/(K_(1)+K_(2))`
`therefore T= 2pi sqrt((m)/(K_("eff")))= 2pi sqrt((m(K_(1)+K_(2)))/(K_(1)K_(2)))`