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(D) The two springs are connected in series. For springs in series,the equivalent force constant $K_{eq}$ is given by:$\frac{1}{K_{eq}} = \frac{1}{K_1

Vedclass · Accepted Answer

$\frac{1}{\pi} \sqrt{\frac{K}{6M}}$

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(D) The two springs are connected in series. For springs in series,the equivalent force constant $K_{eq}$ is given by:$\frac{1}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2}$Given $K_1 = K$ and $K_2 = 2K$,we have:$\frac{1}{K_{eq}} = \frac{1}{K} + \frac{1}{2K} = \frac{2+1}{2K} = \frac{3}{2K}$Therefore,$K_{eq} = \frac{2K}{3}$.The frequency of oscillation $f$ for a spring-mass system is given by:$f = \frac{1}{2\pi} \sqrt{\frac{K_{eq}}{M}}$Substituting the value of $K_{eq}$:$f = \frac{1}{2\pi} \sqrt{\frac{2K/3}{M}} = \frac{1}{2\pi} \sqrt{\frac{2K}{3M}}$Wait,re-evaluating the calculation: $f = \frac{1}{2\pi} \sqrt{\frac{2K}{3M}}$. Let's check the options. Option $D$ is $\frac{1}{\pi} \sqrt{\frac{K}{6M}}$.Note that $\frac{1}{2\pi} \sqrt{\frac{2K}{3M}} = \frac{1}{\pi} \sqrt{\frac{2K}{4 \cdot 3M}} = \frac{1}{\pi} \sqrt{\frac{2K}{12M}} = \frac{1}{\pi} \sqrt{\frac{K}{6M}}$.Thus,option $D$ is correct.

Two springs of force constant $K$ and $2K$ are connected in series to a mass $M$ as shown in the figure. The frequency of oscillation of the mass is

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