One end of a long metallic wire of length L₀ is tied to the ceiling. The other end is tied to a massless spring of spring constant k. A mass m hangs freely from the free end of the spring. The area of cross-section and the 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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$π \sqrt{\frac{m (YA + {kL}_{0})}{kYA}}$

$2 π \sqrt{\frac{mYA}{{kL}_{0}}}$

$2 π \sqrt{\frac{m}{k}}$

$2 π \sqrt{\frac{m (YA + {kL}_{0})}{kYA}}$

answer is D.

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Detailed Solution

For the wire, $Y = \frac{{FL}_{0}}{{AL}_{0}} \Rightarrow \frac{F}{{ΔL}_{0}} = \frac{YA}{L_{0}} ∴$ For the wire,
Spring and wire are in series $k_{eq} = \frac{kk'}{k + k'} = \frac{kYA}{L_{0} (k + \frac{YA}{L_{0}})}$
or $k_{eq} = \frac{kYA}{YA + {kL}_{0}}, T = 2 π \sqrt{\frac{m}{k_{eq}}} = 2 π \sqrt{\frac{m (YA + {kL}_{0})}{kYA}}$

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