Question

A uniform wire of length $L$ and diameter $D$ and density $\rho $ is stretched under a tension $T$ . The correct relation between its fundamental frequency $f,$ the length $L$ and the diameter $D$ is

• A. $f \propto \frac{1}{L D^{2}}$
• B. $f \propto \frac{1}{D^{2}}$
• C. $f \propto \frac{1}{L D}$
• D. $f \propto \frac{1}{L \sqrt{D}}$

Answer 1

Answer: (C)

Since, we know,
Frequency, $f=\frac{n}{2 L} \, \sqrt{\frac{T}{m}}$
Where, $T=$ tension in the string,
$L=$ Length of the string
And $m=$ linear mass density of the string.
$\therefore m=\frac{M}{L}=\frac{\pi \left(\frac{D}{2}\right)^{2} \cdot L \cdot \rho }{L}=\frac{\pi D^{2} \rho }{4}$
Thus, $f=\frac{1}{2 L}\cdot \sqrt{\frac{T}{\frac{\pi D^{2} \rho }{4}}}$
For fundamental frequency, $n=1$
$f=\frac{1}{2 L \frac{D}{2}} \, \sqrt{\frac{T}{\pi ​ \rho }}=\frac{1}{L D}\cdot \sqrt{\frac{T}{\pi ​ \rho }}$
$f \propto \frac{1}{L D}$