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# A mass m is suspended at the end of a massless wire of length L and cross – sectional area A. If Y is the Young’s modulus of the material of the wire, the frequency of oscillations along the vertical line is given by

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## The correct option is D Let l be the increase in the length of the wire due to the force F = m g (see Fig. 9.31). Then Stress=ForceArea=FA Strain=change in lengthOriginal length=lL By definition, Yoiung’s modulus is Y=StressStrain=mgLlA or F=mg=YAlL This is the force acting upwards in the equilibrium state. If the mass is pulled down a little through a distance x, so that the total extension in the string is (l + x ), then the force in the wire acting upwards will be =YAL(l+x) And downward force is F = m g. The restoring force is the net downward force. Hence, Restoring force =mg−YAL(l+x) =YAlL−YAL(l+x)=−YAxL ∴ Acceleration (a) of the mass = forcemass =YAxL=−ω2x Where ω=√YAmL is the angular frequency of the resulting motion which is simple harmonic. Thus, Frequency of oscillation (n) = ω2π=12π√YAmL  Suggest Corrections  4      Similar questions  Related Videos   Applying SHM
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