A particle at the end of a spring executes S.H,M with a period `t_(2)` If the period of oscillation with two spring in .

Options

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

Answer

B

Explanation

Time period of spring `T = 2pi sqrt(((m)/(k)))`
`k` being the force constant of spring.
For first spring. `t_(1) = 2pi sqrt(((m)/(k_(1)))) ….(1)`
For second spring `t_(2) = 2pi sqrt(((m)/(k_(2)))) …(2)`
The effective force constant in their series combination is `k = (k_(1)k_(2))/(k_(1)+k_(2))`. Therefore, time period of combination `T = 2pi sqrt([(m(k_(1)+k_(2)))/(k_(1)k_(2))])`
`rArr T^(2) = (4pi^(2)m(k_(1)+k_(2)))/(k_(1)k_(2)) ....(3)`
From equations (1) and (2), we obtain
`t_(1)^(2) +t_(2)^(2) =4pi^(2) ((m)/(k_(1))+(m)/(k_(2))) :. t_(1)^(2) +t_(2)^(2) =T^(2)` [from eq (3)]