Q.
A wire of density 9 × 10-3 kg cm-3 is stretched between two clamps 1m apart. The resulting strain in the wire is 4.9 × 10-4. The lowest frequency of the transverse vibrations in the wire is (Young’s modulus of wire Y = 9 × 1010 Nm-2), (to the nearest integer), ______
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answer is 0035.00.
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Detailed Solution
Lowest frequency is fundamental frequencyf0=12lTμ Y=TlAΔl⇒f0=12lTρA TA=YΔll⇒f0=12lYΔlρlGiven : l=1 ; Δll=4.9×10−4; Y=9×1010 ; ρ=9×10+3kgm3So ⇒f0=129×1010×4.9×10−49×103=124.9×103=1249×102=702=35 Hz
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