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 , k2 = 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 $\large T = 2 \pi \sqrt{\frac{m}{k_{eq}}}$
$\large T = 2 \pi \sqrt{\frac{m(Y A + k L)}{Y A k}}$