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A mass m is attached to four springs of spring constants 2.k,2k,-k and k as shown in fig. (2). The mass

s capable of oscillating on friction less horizontal floor. If it is displaced slightly and released then the frequency of resulting S.H.M. would be

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$\frac{1}{2 π} \sqrt{(\frac{3 k}{m})}$

$\frac{1}{2 π} \sqrt{(\frac{2 k}{3 m})}$

$\frac{1}{2 π} \sqrt{(\frac{11 k}{2 m})}$

$\frac{1}{2 π} \sqrt{(\frac{4 k}{m})}$

answer is C.

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Detailed Solution

The resultant force constant is given by
$k^{'} = [\frac{(2 k) (2 k)}{2 k + 2 k}] + (k + k) = 3 k$
We know that
$T = 2 π \sqrt{(\frac{m}{k^{'}})} = 2 π \sqrt{(\frac{m}{3 k})} Now n = \frac{1}{T} = \frac{1}{2 π} \sqrt{(\frac{3 k}{m})}$

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