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Question

# The length of the wire shown in figure (15-E8) between the pulley is 1⋅5 m and its mass is 12⋅0 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.

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Solution

## Given: Length of the wire between two pulleys (L) = 1.5 m Mass of the wire = 12 gm $\mathrm{Mass}\mathrm{per}\mathrm{unit}length,m=\frac{12}{1.5}\mathrm{g}/\mathrm{m}\phantom{\rule{0ex}{0ex}}=8×{10}^{-3}\mathrm{kg}/\mathrm{m}$ $\mathrm{Tension}\mathrm{in}\mathrm{the}wire,T=9×g\phantom{\rule{0ex}{0ex}}=90\mathrm{N}$ Fundamental frequency is given by: ${f}_{0}=\frac{1}{2L}\sqrt{\left(\frac{T}{m}\right)}$ For second harmonic (when two loops are produced): ${f}_{1}=2{f}_{0}=\frac{1}{1.5}\sqrt{\left(\frac{90}{8}×{10}^{-3}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(106.06\right)}{1.5}\phantom{\rule{0ex}{0ex}}=70.7\mathrm{Hz}\approx 70\mathrm{Hz}$

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