One end of a long metallic wire of length L,a | vedclass

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(D) The metallic wire acts as a spring with an effective spring constant $k_1$. From Hooke's law,the restoring force $F$ for an extension $x$ is given

Vedclass · Accepted Answer

$2 \pi \sqrt{\frac{m(k L+Y A)}{k Y A}}$

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(D) The metallic wire acts as a spring with an effective spring constant $k_1$. From Hooke's law,the restoring force $F$ for an extension $x$ is given by $F = \frac{YA}{L}x$. Thus,the spring constant of the wire is $k_1 = \frac{YA}{L}$.Since the wire and the spring are connected in series,the equivalent spring constant $k_{eq}$ of the system is given by $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k}$.Substituting $k_1 = \frac{YA}{L}$,we get $\frac{1}{k_{eq}} = \frac{L}{YA} + \frac{1}{k} = \frac{kL + YA}{kYA}$.Therefore,$k_{eq} = \frac{kYA}{kL + YA}$.The time period of oscillation $T$ for a mass $m$ attached to a spring system is $T = 2\pi \sqrt{\frac{m}{k_{eq}}}$.Substituting the value of $k_{eq}$,we get $T = 2\pi \sqrt{\frac{m(kL + YA)}{kYA}}$.

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