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. $2\pi \sqrt{\frac{m\left(YA+kL\right)}{YAk}}$
• C. $2\pi \left[{\left(\frac{mYA}{kL}\right)}^{1/2}\right]$
• D. $2\pi \left[{\left(\frac{mL}{YA}\right)}^{1/2}\right]$

Answer 1

Answer: (B)

Equivalent force constant for a wire is given by $k=\frac{YA}{L}$ . Because in case of a wire, $F=\frac{YA}{L}\Delta L$ and in case of spring, $F=k\Delta x$ . Comparing these two, we find k of wire $=\frac{YA}{L}$
$k_{\text{eq}}=\frac{k_1{k}_2}{k_1+k_2}=\frac{\frac{YA}{L}k}{\frac{YA}{L}+k}=\frac{YAk}{YA+Lk}$

$\therefore T=2\pi \sqrt{\frac{m}{k_{\text{eq}}}}=2\pi \sqrt{\frac{m\left(YA+Lk\right)}{YAk}}$