A block of mass m, when attached to a uniform ideal spring with force constant k and free length L executes SHM. The spring is then cut in two pieces, one with free length n L and other with free length (1-n) L (n $< 0.5)$ . The block is also divided in the same fraction. The smaller part of the block attached to longer part of the spring executes SHM with frequency $f_{1}$ . The bigger part of the block attached to smaller part of the spring executes SHM with frequency $f_{2}$ . The ratio $f_{1} / f_{2}$ is

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1-n

$\frac{n}{1 - n}$

$\frac{1 + n}{n}$

$\frac{n}{1 + n}$

answer is A.

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Detailed Solution

$T = 2 π \sqrt{\frac{m}{k}} = \frac{1}{f} K_{s} = F o r c e c o n s \tan t o f t h e s m a l l e r p a r t = \frac{(1 + n) K}{n} K_{L} = F o r c e c o n s \tan t o f t h e l a r g e r p a r t = (1 + n) K S i m i l a r l y m_{s} = s m a l l e r m a s s = n m a n d m_{L} = (1 - n) m T h e r e f o r e f_{1} = \frac{1}{2 π} \sqrt{\frac{(1 + n) K}{n m}} a n d f_{2} = \frac{1}{2 π} \sqrt{\frac{(1 + n) K}{n (1 - n) m}} S o \frac{f_{1}}{f_{2}} = \sqrt{(1 - n)}$