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In arrangement given in figure, if the block of mass $m$ is displaced, the frequency is given by
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The correct answer is:
$n=\frac{1}{2 \pi} \sqrt{\left(\frac{k_1+k_2}{m}\right)}$
With respect to the block the springs are connected in parallel combination.
$\therefore$ Combined stiffness $k=k_1+k_2$ and $n=\frac{1}{2 \pi} \sqrt{\frac{k_1+k_2}{m}}$
$\therefore$ Combined stiffness $k=k_1+k_2$ and $n=\frac{1}{2 \pi} \sqrt{\frac{k_1+k_2}{m}}$
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