Question 119 of 194

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

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Answer: 2

In figure (a), the springs are connected in series. Therefore,
${K}_{eq}=\dfrac{{K}_{1}{K}_{2}}{{K}_{1}+{K}_{2}}=\dfrac{K\times 2K}{K+2K}=\dfrac{2K}{3}$
Therefore, the time period
${T}_{a}=2\pi \sqrt{\dfrac{m}{{K}_{eq}}}=2\pi \sqrt{\dfrac{m}{\dfrac{2K}{3}}}=2\pi \sqrt{\dfrac{3m}{2K}}$
In figure (b), the springs are connected in parallel. Therefore,
${K}_{eq}={K}_{1}+{K}_{2}=3K$
Therefore, the time period
${T}_{b}=2\pi \sqrt{\dfrac{m}{{K}_{eq}}}=2\pi \sqrt{\dfrac{m}{3K}}$
Now,
$\dfrac{{T}_{a}}{{T}_{b}}=\sqrt{\dfrac{9}{2}}\Rightarrow {T}_{b}=\dfrac{\sqrt{2}}{3}{T}_{a}=\sqrt{2}s$
Hence, $x=2$