Periodic time of oscillation $T_{1}$ is obtained when a mass is suspended from a spring. If another spring is used with same mass then periodic time of oscillation is $T_{2}$ . Now if this mass is suspended from series combination of above springs then calculate the time period.

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$T_{e q} = 4 \sqrt{T_{1}^{2} + T_{2}^{2}}$

$T_{e q} = 3 \sqrt{T_{1}^{2} + T_{2}^{2}}$

$T_{e q} = 2 \sqrt{T_{1}^{2} + T_{2}^{2}}$

$T_{e q} = \sqrt{T_{1}^{2} + T_{2}^{2}}$

answer is D.

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Detailed Solution

$T_{1} = 2 π \sqrt{\frac{m}{k_{1}}} \Rightarrow k_{1} = \frac{4 π^{2} m}{T_{1}^{2}}$ and $T_{2} = 2 π \sqrt{\frac{m}{k_{2}}} \Rightarrow k_{2} = \frac{4 π^{2} m}{T_{2}^{2}}$ so $K_{e q} = \frac{4 π^{2} m}{T_{e q}^{2}}$

Inseries combination $\frac{1}{K} = \frac{1}{K_{1}} + \frac{1}{K_{2}} \Rightarrow \frac{T_{e q}^{2}}{4 π^{2} m} = \frac{T_{1}^{2}}{4 π^{2} m} + \frac{T_{2}^{2}}{4 π^{2} m} \Rightarrow T_{e q} = \sqrt{T_{1}^{2} + T_{2}^{2}}$