A wire of density 9×103 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)
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a
40 Hz
b
35 Hz
c
30 Hz
d
25 Hz
answer is B.
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Detailed Solution
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=35Hz