A mass `m` is suspended separately by two different spring of spring constant `K_(1)` and `k_(2)` given the time period `t_(1)` and`t_(2)` respectively if the same mass `m` is shown in the figure then time period `t` is given by the relation

Options

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

Answer

D

Explanation

`t_(1)=2pi sqrt((m)/(K_(1))) therefore t_(1)^(2)= 4pi^(2) (m)/(K_(1))`
`therefore K_(1)=(4pi_(m)^(2))/(t_(1)^(2))`
and `t_(2)=2pi sqrt((m)/(K_(2)) therefore t_(2)^(2)=4pi^(2)(m)/(K_(2))`
`therefore K_(2)=(4pi_(m)^(2))/(t_(2)^(2))`
and when they are joined in parallel, `K=K_(1)+K_(2)`
and `t=2pi sqrt((m)/(K_(1)+K_(2)) therefore t^(2)=4pi^(2)(m)/((K_(1)+K_(2)))`
and `K_(1)+K_(2)=(4pi^(2)m)/(t^(2))`
`therefore` from (1), (2) and (3),
`(4pi^(2)m)/(t^(2))=(4pi^(2)m)/(t_(1)^(2))+(4pi^(2)m)/(t_(2)^(2))`
`therefore t^(-2)=t_(1)^(-2)+t_(2)^(-2)`