wave_mechanics critical_thinking Question 32
Question: 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)
[UPSEAT 2000; Pb. PET 2004]
Options:
A) 40 Hz
B) 35 Hz
C) 30 Hz
D) 25 Hz
Show Answer
Answer:
Correct Answer: B
Solution:
For wire if M = mass, r = density, A = Area of cross section V = volume, l = length, Dl = change in length Then mass per unit length $ m=\frac{M}{l}=\frac{Al\rho }{l}=A\rho $ And Young’s modules of elasticity $ y=\frac{T/A}{\Delta l/l} $
Therefore $ T=\frac{Y\Delta lA}{l} $ .
Hence lowest frequency of vibration $ n=\frac{1}{2l}\sqrt{\frac{T}{m}} $
$ =\frac{1}{2l}\sqrt{\frac{y( \frac{\Delta l}{l} )A}{A\rho }}=\frac{1}{2l}\sqrt{\frac{y\Delta l}{l\rho }} $
Therefore $ n=\frac{1}{2\times 1}\sqrt{\frac{9\times 10^{10}\times 4.9\times {{10}^{-4}}}{1\times 9\times 10^{3}}}=35Hz $
BETA
