One end of a long metallic wire of length $L$, area of cross-section $A$ and Young's modulus $Y$ is tied to…
- $2 \pi \sqrt{\frac{m}{k}}$
- $2 \pi \sqrt{\frac{m Y A}{k L}}$
- $2 \pi \sqrt{\frac{m(k A+Y L)}{k Y A}}$
- $2 \pi \sqrt{\frac{m(k L+Y A)}{k Y A}}$
Solution

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, $\quad T=2 \pi \sqrt{\frac{m}{k_{e q}}} \Rightarrow T=2 \pi \sqrt{\frac{m(k L+Y A)}{k Y A}}$

Asked in: AP EAMCET 2018 (23 Apr Shift 2)