Question

One end of a long metallic wire of length $L$ 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 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

• A. $2\pi \left(\frac{m}{k}\right)^{1 / 2}$
• B. $2\pi \sqrt{\frac{m \left(Y A + k L\right)}{Y A k}}$
• C. $2\pi \left[\left(\frac{m Y A}{k L}\right)^{1/2}\right]$
• D. $2\pi \left[\left(\frac{m L}{Y A}\right)^{1/2}\right]$

Answer 1

Answer: (B)

Equivalent force constant for a wire is given by $k=\frac{Y A}{L}$ . Because in case of a wire, $F=\frac{Y A}{L}\Delta L$ and in case of spring, $F=k\Delta x$ . Comparing these two, we find k of wire $=\frac{Y A}{L}$
$k_{\text{eq}}=\frac{k_{1} k_{2}}{k_{1} + k_{2}}=\frac{\frac{Y A}{L} k}{\frac{Y A}{L} + k}=\frac{Y A k}{Y A + L k}$

$\therefore T=2\pi \sqrt{\frac{m}{k_{\text{eq}}}}=2\pi \sqrt{\frac{m \left(Y A + L k\right)}{Y A k}}$