In arrangement given in fig. (8), if the block of mass m is displaced, the frequency is given by

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$n = \frac{1}{2 π} \sqrt{(\frac{m}{k_{1} + k_{2}})}$

$n = \frac{1}{2 π} \sqrt{(\frac{k_{1} + k_{2}}{m})}$

$n = \frac{1}{2 π} \sqrt{(\frac{m}{k_{1} - k_{2}})}$

$n = \frac{1}{2 π} \sqrt{(\frac{k_{1} - k_{2}}{m})}$

answer is B.

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Detailed Solution

Let the extension in spring A is x, then the compression in string B will also be x. The restoring force on mass is a force given by .k, x. This force acts towards left' Again a force k₂, x pushes the mass to the left due to compression of spring B. The net restoring force acting on the mass is $(k_{1} + k_{2}) x$
$force constant = (k_{1} + k_{2}) Now T = 2 π \sqrt{(\frac{m}{k_{1} + k_{2}})} Frequency n = \frac{1}{T} = \frac{1}{2 π} \sqrt{\{\frac{(k_{1} + k_{2})}{m}\}}$