Question

A wire of density $\rho$ is stretched between two clamps a distance $L$ apart, while being subjected to an extension $l$. Y is the Young's modulus of the material of the wire. The lowest frequency of transverse vibrations of the wire is given by

• A. $v=\frac{1}{2L}\sqrt{\frac{YL}{l\rho }}$
• B. $v=\frac{1}{2L}\sqrt{\frac{Y\rho L}{l^2}}$
• C. $v=\frac{1}{2L}\sqrt{\frac{Yl}{L\rho }}$
• D. $v=\frac{1}{2L}\sqrt{\frac{L\rho }{Yl}}$

Answer 1

The correct option is C $v=\frac{1}{2L}\sqrt{\frac{Yl}{L\rho }}$

$Y=\frac{Stress}{Strain}=\frac{\frac{T}{A}}{\frac{l}{L}}=\frac{TL}{Al}$ where A is the area of cross-section of the wire.
Thus, $T=\frac{YAl}{L}$ .... (i)
Now, mass per length of the wire, $\mu =\frac{Mass}{Length}=\frac{Volume\times Density}{Length}$
$=Area\times Density=A\rho$ ..... (ii)
The lowest frequency (i.e. frequency of the fundamental mode) is given by
$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$
Using (i) and (ii), we get $f=\frac{1}{2L}\sqrt{\frac{Yl}{L\rho }}$
Hence, the correct choice is (c).