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. $f\propto \frac{1}{LD}$
• D. $f\propto \frac{1}{L\sqrt{D}}$

Answer 1

Answer: (C)

Since, we know,
Frequency, $f=\frac{n}{2L}\sqrt{\frac{T}{m}}$
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, $f=\frac{1}{2L}\cdot \sqrt{\frac{T}{\frac{\pi D^2\rho }{4}}}$
For fundamental frequency, $n=1$
$f=\frac{1}{2L\frac{D}{2}}\sqrt{\frac{T}{\pi \rho }}=\frac{1}{LD}\cdot \sqrt{\frac{T}{\pi \rho }}$
$f\propto \frac{1}{LD}$