One end of a long metallic wire of length L tied to the ceiling....

Calculation of spring constant of wire Young's modulus of wire, (\mathrm{Y}=\frac{\Delta \mathrm{F} / \mathrm{A}}{\Delta \mathrm{L} / \mathrm{L}}=\frac{\Delta \mathrm{F}}{\Delta \mathrm{L}} \times \frac{\mathrm{L}}{\mathrm{A}}) Due to elasticity of the wire, it behaves as equivalent spring (\left(\mathrm{K}_{\text{spring }}=\mathrm{F}/ \Delta \mathrm{x}\right)). Hence, Spring constant for Wire, (\mathrm{K}_{\mathrm{w}}=\frac{\Delta \mathrm{F}}{\Delta \mathrm{L}}=\mathrm{Y} \frac{\mathrm{A}}{\mathrm{L}}$) Calculation of Equivalent spring constant Let a force (\mathrm{F}$) be applied to the end of the spring. The wire will extend. Total extension (\Delta $\mathrm{S}=\mathrm{x}$_{\mathrm{wire}}+\mathrm{x}_{\text{spring }}=\frac{\mathrm{F}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{w}}}+\frac{\mathrm{F}_{\mathrm{S}}}{\mathrm{K}}$) But (\mathrm{F}_{\mathrm{S}}=\mathrm{F}_{\mathrm{w}} \quad$) (By Newton's, third Law, Both will apply equal and opposite forces at their joint A ) (\therefore $\Delta$\mathrm{S}=\mathrm{F}$_{\mathrm{w}} \left(\frac{1}{\mathrm{ K}_{\mathrm{w}}}+\frac{1}{\mathrm{ K}}\right)$) So, Equivalent spring constant \$( \left(\mathrm{K}_{\text{eq }}\right)$) is given by: (\mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{F}_{\mathrm{eq}}}{\mathrm{x}_{\mathrm{eq}}}=\frac{\mathrm{F}_{\mathrm{w}}}{\Delta \mathrm{S}}=\frac{1}{\frac{1}{\mathrm{ K}_{\mathrm{w}}}+\frac{1}{\mathrm{ K}}}) \$(\mathrm{K}_{\mathrm{eq}}=\frac{(\mathrm{K})\left(\mathrm{K}_{\mathrm{w}}\right)}{\mathrm{K}_{\mathrm{w}}+\mathrm{K}}=\frac{\mathrm{KY} \frac{\mathrm{A}}{\mathrm{L}}}{\mathrm{Y} \frac{\mathrm{A}}{\mathrm{L}}+\mathrm{K}}=\frac{\mathrm{KYA}}{\mathrm{YA}+\mathrm{KL}}$) Calculation of Time period Angular frequency (\omega) for spring block system is given by: \$(\$omega=\sqrt{\frac{\mathrm{K}_{\text{eq }}}{\mathrm{m}}}=\sqrt{\frac{\mathrm{KYA}}{\mathrm{m}(\mathrm{YA}+\mathrm{KL})}}$) (\therefore) Time period, (\mathrm{T}=\frac{2 \pi}{\omega}=$2 \pi$\sqrt{\frac{(\mathrm{YA}+\mathrm{KL}) \mathrm{m}}{\mathrm{KYA}}}$) Hence, Option B is correct