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

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Correct answer: B. 2π (m (YA + kL))/(YAk)

Equivalent force constant for a wire is given by $k = \dfrac{YA}{L}$ . Because in case of a wire, $F = \dfrac{YA}{L} ∆ L$ and in case of spring, $F = k ∆ x$ . Comparing these two, we find k of wire $= \dfrac{YA}{L}$

${k}_{eq} = \dfrac{{k}_{1}{k}_{2}}{{k}_{1} + {k}_{2}} = \dfrac{\dfrac{YA}{L}k}{\dfrac{YA}{L} + k} = \dfrac{YAk}{YA + Lk}$

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

$∴ T = 2\pi\sqrt{\dfrac{m}{k\text{eq}}} = 2\pi\sqrt{\dfrac{m\left(YA + Lk\right)}{YAk}}$
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One end of a long metallic wire of length L is tied to the ceiling....

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