A steel wire of length 1 m and mass 0.1 kg | vedclass

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(A) The thermal strain produced due to a change in temperature $\Delta t$ is given by $	ext{Strain} = \alpha \Delta t$.Using Hooke's Law,the thermal

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$11$

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(A) The thermal strain produced due to a change in temperature $\Delta t$ is given by $	ext{Strain} = \alpha \Delta t$.Using Hooke's Law,the thermal stress is $	ext{Stress} = Y 	imes 	ext{Strain} = Y \alpha \Delta t$.The tension $T$ in the wire is $T = 	ext{Area} 	imes 	ext{Stress} = A Y \alpha \Delta t$.Given $A = 10^{-6} \ m^2$,$Y = 2 	imes 10^{11} \ N/m^2$,$\alpha = 1.21 	imes 10^{-5} /^{\circ} C$,and $\Delta t = 20^{\circ} C$,we have:$T = 10^{-6} 	imes 2 	imes 10^{11} 	imes 1.21 	imes 10^{-5} 	imes 20 = 48.4 \ N$.The linear mass density $\mu = \frac{m}{\ell} = \frac{0.1 \ kg}{1 \ m} = 0.1 \ kg/m$.The fundamental frequency $f$ is given by $f = \frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}$.Substituting the values: $f = \frac{1}{2 	imes 1} \sqrt{\frac{48.4}{0.1}} = \frac{1}{2} \sqrt{484} = \frac{22}{2} = 11 \ Hz$.

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