$(a)\;\large\frac{1}{12}\;\sqrt{\large\frac{K}{m}}\qquad(b)\;\large\frac{1}{3}\;\sqrt{\large\frac{K}{m}}\qquad(c)\;\large\frac{1}{6}\;\sqrt{\large\frac{K}{m}}\qquad(d)\;None\;of\;these$

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Answer : (c) $\;\large\frac{1}{6}\;\sqrt{\large\frac{K}{m}}$

Explanation :

Angular frequency dependent only on the mass and spring constant as $\;w=\sqrt{\large\frac{K}{m}}$

Therefore , The total mass that sticks is

$m+2m+-------+\theta m=36 m \qquad \; $ (inelostic collision)

$w=\sqrt{\large\frac{K}{36m}}=\large\frac{1}{6}\;\sqrt{\large\frac{K}{m}}\;.$

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