Waves

JEE Main level, numerical answer, medium

65 of 142

Question 65 of 142

A wire of density 8×103kgm−38\times {10}^{3}kg{m}^{-3} is stretched between two clamps 0.5m0.5m apart. The extension developed in the wire is 3.2×10−4m3.2\times {10}^{-4}m. If Y=8×1010Nm−2Y=8\times {10}^{10}N{m}^{-2}, the fundamental frequency of vibration in the wire will be _____ HzHz

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Answer: 80

Using the relation of Young's modulus,
TA=Y(ΔLL)\dfrac{T}{A}=Y\left(\dfrac{\Delta L}{L}\right)
⇒T=(YΔLL×A)\Rightarrow T=\left(\dfrac{Y\Delta L}{L}\times A\right)
The linear mass density is μ=(mL)\mu =\left(\dfrac{m}{L}\right).
So,
Tμ=YΔLAL(mL)=Y(ΔL)×LAL(m)=(YΔLL)×(1ρ)\dfrac{T}{\mu }=\dfrac{Y\Delta LA}{L\left(\dfrac{m}{L}\right)}=\dfrac{Y(\Delta L)\times LA}{L(m)}=\left(\dfrac{Y\Delta L}{L}\right)\times \left(\dfrac{1}{\rho }\right)
Substituting the values,
Tμ=8×1010×3.2×10−40.5×(18×103)=6.4×103\dfrac{T}{\mu }=\dfrac{8\times {10}^{10}\times 3.2\times {10}^{-4}}{0.5}\times \left(\dfrac{1}{8\times {10}^{3}}\right)=6.4\times {10}^{3}
⇒Tμ=64×102\Rightarrow \dfrac{T}{\mu }=64\times {10}^{2}
The fundamental frequency is given by f=12LTμf=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu }}.
⇒Tμ=8×10=80ms−1\Rightarrow \sqrt{\dfrac{T}{\mu }}=8\times 10=80m{s}^{-1}
Therefore,
f=(801)=80Hzf=\left(\dfrac{80}{1}\right)=80Hz

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