Q: 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. A mass m hangs freely from the free end of the spring. It is slightly pulled down and released. Its time period is given by

(a) $ \displaystyle 2\pi \sqrt{\frac{m}{k}}$

(b) $ \displaystyle 2\pi \sqrt{\frac{mYA}{k L}}$

(c) $ \displaystyle 2\pi \sqrt{\frac{mY}{k}}$

(d) $ \displaystyle 2\pi \sqrt{\frac{m(kL+YA)}{kYA}}$

Ans:(d)

Sol: For wire ,

$ \displaystyle Y = \frac{F L}{A \Delta L}$

$ \displaystyle F = (\frac{A Y}{L} )\Delta L $

$ \displaystyle F = K’ \Delta L $

Where $ \displaystyle K’ = \frac{A Y}{L}$

For Spring , Force Constant = K

Both Wire & Spring are in Series ,

$ \displaystyle \frac{1}{K_{eq}} = \frac{1}{K} + \frac{1}{K’} $

$\displaystyle \frac{1}{K_{eq}} = \frac{1}{K} + \frac{L}{A Y} $

Time Period , $ \displaystyle T = 2\pi \sqrt{\frac{m}{K_{eq}}} $

$ \displaystyle T = 2\pi \sqrt{m(\frac{1}{K} + \frac{L}{A Y})} $

$ \displaystyle T = 2\pi \sqrt{\frac{m(YA + KL)}{kYA}}$