Q. One end of a long metallic wire of length $L$ is tied to the ceiling. The other end is tied to 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) $2\pi\left( \frac{m}{K} \right)$
B) $2\pi\left\{ \frac{(YA + KL)m}{YAK} \right\}^{1/2}$
C) $2\pi\frac{mYA}{KL}$
D) $2\pi\frac{mL}{YA}$
Explanation
(b) The wire may be treated as a spring for which force constant
$k_{1} = \frac{Force}{Extension} = \frac{YA}{L}\ \left( \because Y = \frac{F}{A} \times \frac{L}{\mathrm{\Delta}L} \right)$
Spring constant of the spring $k_{2} = K$
Hence spring constant of the combination (series)
$k_{eq} = \frac{k_{1}k_{2}}{k_{1} + k_{2}} = \frac{(YA/L)K}{(YA/L) + K} = \frac{YAK}{YA + KL}$
$\therefore$ Time period $T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\left\lbrack \frac{(YA + KL)m}{YAK} \right\rbrack^{1/2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.