Questions
A wire of density 9103 kg /m3 is stretched between two clamps 1 m apart and is subjected to an extension of 4.9 10-4 m. The lowest frequency of transverse vibration in the wire is (Y = 9 1010 N / m2)
detailed solution
Correct option is B
For wire if M = mass, ρ = density, A = Area of cross section V = volume, l = length, Δl = change in length Then mass per unit length m=Ml=Alρl=AρAnd Young’s modules of elasticity y=T/AΔl/l⇒T=YΔlAl. Hence lowest frequency of vibration n=12lTm=12lyΔllAAρ=12lyΔllρ⇒n=12×19×1010×4.9×10−41×9×103=35HzSimilar Questions
A wire having a linear density of 5 g/m is stretched between two rigid supports with a tension of 450 N. It is observed that the wire resonates at a frequency of 420 Hz. The next highest frequency at which the same wire resonates is 490 Hz. Find the length of the wire. (in m)
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