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(B) The frequency of oscillation for a mass $m$ connected to a spring of constant $K$ is given by $n = \frac{1}{2\pi} \sqrt{\frac{K}{m}}$.For springs

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$\sqrt{n_1^2 + n_2^2}$

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(B) The frequency of oscillation for a mass $m$ connected to a spring of constant $K$ is given by $n = \frac{1}{2\pi} \sqrt{\frac{K}{m}}$.For springs $S_1$ and $S_2$ with spring constants $K_1$ and $K_2$ respectively,we have:$n_1 = \frac{1}{2\pi} \sqrt{\frac{K_1}{m}} \implies n_1^2 = \frac{1}{4\pi^2} \frac{K_1}{m} \implies K_1 = 4\pi^2 m n_1^2$$n_2 = \frac{1}{2\pi} \sqrt{\frac{K_2}{m}} \implies n_2^2 = \frac{1}{4\pi^2} \frac{K_2}{m} \implies K_2 = 4\pi^2 m n_2^2$In the given figure,the mass $m$ is connected to two springs in parallel. The effective spring constant $K_{eff}$ is $K_1 + K_2$.The new frequency $n_{eff}$ is given by:$n_{eff} = \frac{1}{2\pi} \sqrt{\frac{K_{eff}}{m}} = \frac{1}{2\pi} \sqrt{\frac{K_1 + K_2}{m}}$Substituting the values of $K_1$ and $K_2$:$n_{eff} = \frac{1}{2\pi} \sqrt{\frac{4\pi^2 m n_1^2 + 4\pi^2 m n_2^2}{m}} = \frac{1}{2\pi} \sqrt{4\pi^2 (n_1^2 + n_2^2)} = \sqrt{n_1^2 + n_2^2}$

When a mass $m$ is connected individually with two springs $S_1$ and $S_2$,the oscillation frequencies are $n_1$ and $n_2$. If the mass $m$ is attached to the springs as shown in the figure,the oscillation frequency would be

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