Two springs of force constant k and 2k are connected to a mass shown in figure. The frequency of oscillation of the mass is :

- \frac{1}{2 π} \sqrt{\frac{m}{k}}

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\frac{1}{2 π} \sqrt{\frac{2 k}{m}}

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\frac{1}{2 π} \sqrt{\frac{3 k}{m}}

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\frac{1}{2 π} \sqrt{\frac{m}{k}}

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Solution

The correct option is C $\frac{1}{2 π} \sqrt{\frac{3 k}{m}}$
Let the mass m be displaced toward right by a distance x. Then, spring 1 will be extended and the spring 2 will be compressed by en amount x. On releasing mass m the restoring force on m due to spring 1 is $F_{1} = - k x$ and restoring force due to spring 2 is $F_{2} = - 2 k x$
Total restoring force on mass m is
$F = F_{1} + F_{2}$
$F = - k x - 2 k x = - 3 k x$
$F = \frac{m d^{2} x}{d t^{2}} = - 3 k x$
$\frac{d^{2} x}{d t^{2}} = - \frac{3 k}{m} x$
$\frac{d^{2} x}{d t^{2}} + \frac{3 k}{m} x = 0$
This is similar to $\frac{d^{2} x}{d t^{2}} + ω^{2} x = 0$
Which respresent equation of SHM. Comparing the two equations, $ω^{2} = \frac{3 k}{m}$
$ω = \sqrt{\frac{3 k}{m}}$
$2 π f = \sqrt{\frac{3 k}{m}}$
$f = \frac{1}{2 π} \sqrt{\frac{3 k}{m}}$

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