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Asked by sarveshvibrantacademy | 05 Apr, 2019, 10:28: AM
answered-by-expert Expert Answer
youngs modulus Y = begin mathsize 12px style fraction numerator T divided by A over denominator increment l divided by l end fraction end style  ................(1)
where T is tension subjected by wire, A is area of cross section of wire, Δl is change in length and l is original length.
 
when the peg is turned for one quarter, Δl = (1/4)π×5×10-3 m = 1.25π×10-3 m
 
from eqn.(1) , we get T as,  T = Y×(Δl/l)×A  = 20×1010 × [ (1.25π×10-3 ) / (6×10-2 ) ]×(π/4)×(8×10-4)2  ≈ 6580 N
 
Linear mass density, μ = mass of unit length of string = (π/4)×(8×10-4)2 × 1 × 7800 = 98×10-5 kg/m
 
velocity of transverse wave, v = begin mathsize 12px style square root of T over mu end root space equals space square root of fraction numerator 6580 over denominator 98 cross times 10 to the power of negative 5 end exponent end fraction end root space equals space 2591 space m divided by s end style
for fundamental mode, wavelength λ = 2l = 2 × 6×10-2 m = 12× 10-2 m 
 
fundamental frequency,  n = v/λ = 2591/ (12× 10-2 )  = 21592 Hz = 21.6 kHz
Answered by Thiyagarajan K | 05 Apr, 2019, 14:15: PM

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