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One end of a long metallic wire of length $L$, area of cross-section $A$ and Young's modulus $Y$ is tied to the ceiling. The other end is tied to a massless spring of force constant $k$ and a mass $m$ is hung from the free end of the spring. If $m$ is slightly pulled down and released, then its time period of oscillation is
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Verified Answer
The correct answer is:
$2 \pi \sqrt{\frac{m(k L+Y A)}{k Y A}}$
For oscillating mass at end of a rod. Restoring force
$$
=\frac{Y A}{L} \cdot x
$$
So, $k_1=$ spring constant for a rod is $\frac{Y A}{L}$.
If a rod and spring are connected, then it is a series combination.
So, $\left(k_{\text {eq }}\right)$
$$
\begin{aligned}
\text { system } & =\frac{k_1 k_2}{k_1+k_2} \\
& =\frac{k Y A / L}{k+\frac{Y A}{L}}=\frac{k Y A}{k L+Y A}
\end{aligned}
$$
So, $\quad T=2 \pi \sqrt{\frac{m}{k_{e q}}} \Rightarrow T=2 \pi \sqrt{\frac{m(k L+Y A)}{k Y A}}$
$$
=\frac{Y A}{L} \cdot x
$$
So, $k_1=$ spring constant for a rod is $\frac{Y A}{L}$.
If a rod and spring are connected, then it is a series combination.
So, $\left(k_{\text {eq }}\right)$
$$
\begin{aligned}
\text { system } & =\frac{k_1 k_2}{k_1+k_2} \\
& =\frac{k Y A / L}{k+\frac{Y A}{L}}=\frac{k Y A}{k L+Y A}
\end{aligned}
$$
So, $\quad T=2 \pi \sqrt{\frac{m}{k_{e q}}} \Rightarrow T=2 \pi \sqrt{\frac{m(k L+Y A)}{k Y A}}$
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