A wire of density $9 \times 10^{3} \text{ kg/m}^{3}$ is stretched between two clamps 1 m apart and is subjected to an extension of $4.9 \times 10^{-4} \text{ m}$ . The lowest frequency of transverse vibration in the wire is $(Y = 9 \times 10^{10} \text{ N}/$ $m^{2}$ )
Text Solution
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For wire if
M = mass, ρ ρ = density, A = Area of cross section
V = volume, l = length, Δ Δ l = change in length
Then mass per unit length $m = \frac{M}{l} = \frac{Al\rho}{l} = A\rho$
And Young’s modules of elasticity $y = \frac{T/A}{\Delta l / l}$
⇒ ⇒ $T = \frac{Y \Delta A}{l}$ . Hence lowest frequency of vibration $n = \frac{1}{2l} \sqrt{\frac{T}{m}} = \frac{1}{2l} \sqrt{\frac{y \left(\frac{\Delta l}{l}\right) A}{A \rho}} = \frac{1}{2l} \sqrt{\frac{y \Delta l}{l \rho}}$
⇒ ⇒ $n = \frac{1}{2 \times 1} \sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{-4}}{1 \times 9 \times 10^{3}}} = 35 \mathrm{Hz}$
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