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

Question

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 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

TestHub · Accepted Answer

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

$k_{eq} = \frac{k_{1} k_{2}}{k_{1} + k_{2}} = \frac{\frac{Y A}{L} k}{\frac{Y A}{L} + k} = \frac{Y A k}{Y A + L k}$

$∴ T = 2 π \sqrt{\frac{m}{k_{eq}}} = 2 π \sqrt{\frac{m (Y A + L k)}{Y A k}}$

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