Q: 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 the 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) $\large 2 \pi \sqrt{\frac{m}{k}}$

(b) $\large 2 \pi \sqrt{\frac{m(Y A + k L)}{Y A k}}$

(c) $\large 2 \pi \sqrt{\frac{m Y A}{k L}}$

(d) $\large 2 \pi \sqrt{\frac{m L}{Y A}}$

Ans: (b)

Sol: For rod , $\large Y = \frac{F L }{A \Delta L}$

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

$\large k_1 = \frac{A Y}{L}$

For Spring , k_{2} = k

Equivalent Force constant $\large k_{eq} = \frac{k_1 k_2}{k_1 + k_2}$

$\large k_{eq} = \frac{(Y A/L)k}{YA/L + k}$

$\large k_{eq} = \frac{Y A k}{Y A + L k} $

Time Period $\large T = 2 \pi \sqrt{\frac{m}{k_{eq}}}$

$\large T = 2 \pi \sqrt{\frac{m(Y A + k L)}{Y A k}}$