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One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to massless spring of spring constant K. A mass (m) hangs freely from the free end of the spring. The area of cross-section and Young's modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released it will oscillate with a time period T equal to

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$2 π \sqrt{\frac{m}{k}}$

$2 π \sqrt{\frac{m (YA + KL)}{YAK}}$

$2 π \sqrt{\frac{mYA}{kL}}$

$2 π \sqrt{\frac{mL}{YA}}$

answer is B.

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Detailed Solution

$Y = \frac{F}{A} \times \frac{L}{e} \Rightarrow \frac{YA}{L} = \frac{F}{e} = K_{1}, K_{2} = K$

As the spring is connected in series with the rod, $K_{eff} = \frac{K_{1} K_{2}}{K_{1} + K_{2}}$

$K_{eff} = \frac{\frac{YA}{L} K .}{\frac{YA}{L} + K} = \frac{\frac{YAK}{L}}{\frac{YA + LK}{L}} = \frac{YAK}{YA + LK}$

$T = 2 π \sqrt{\frac{m}{K_{eff}}} \Rightarrow T = 2 π \sqrt{\frac{m (YA + LK)}{YAK}}$

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