As per given figures, two springs of spring constants $K$ and $2 K$ are connected to mass $m$ . If the period of oscillation in figure (a) is $3 s$ , then the period of oscillation in figure (b) will be $\sqrt{x} s$ . The value of $x$ is _____ .

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TestHub · Accepted Answer

In figure (a), the springs are connected in series. Therefore,
$K_{eq} = \frac{K_{1} K_{2}}{K_{1} + K_{2}} = \frac{K \times 2 K}{K + 2 K} = \frac{2 K}{3}$
Therefore, the time period
$T_{a} = 2 π \sqrt{\frac{m}{K_{eq}}} = 2 π \sqrt{\frac{m}{\frac{2 K}{3}}} = 2 π \sqrt{\frac{3 m}{2 K}}$
In figure (b), the springs are connected in parallel. Therefore,
$K_{e q} = K_{1} + K_{2} = 3 K$
Therefore, the time period
$T_{b} = 2 π \sqrt{\frac{m}{K_{eq}}} = 2 π \sqrt{\frac{m}{3 K}}$
Now,
$\frac{T_{a}}{T_{b}} = \sqrt{\frac{9}{2}} \Rightarrow T_{b} = \frac{\sqrt{2}}{3} T_{a} = \sqrt{2} s$
Hence, $x = 2$

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