Question

A uniform wire of length,$L$, 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. $f\propto \frac{1}{LD}$
• B. $f\propto \frac{1}{L\sqrt{D}}$
• C. $f\propto \frac{1}{D^2}$
• D. $f\propto \frac{1}{L{D}^2}$

Answer 1

The correct option is A

$f\propto \frac{1}{LD}$

Step 1. Given Data,

Length,$L$

Diameter, $D$

Density, $\rho$

Tension, $T$

Fundamental Frequency, $f$

Step 2. Calculation of mass per unit length $\mu$,

The wire is in the form of a cylinder, so its volume is

$V=\pi r^{2}L$ (where $r$ is the radius.)

$V=\frac{\pi D^{2}L}{4}$(since radius $r=\frac{D}{2}$)

Density,

$$\begin{gathered} \rho =\frac{m}{V} \\ \rho =\frac{4m}{\pi D^{2}L} \end{gathered}$$ [Density, $\rho$ is equal to the mass, $m$ per unit volume, $V$.]

Therefore, mass is

$m=\frac{\rho \pi L{D}^2}{4}$

Then, mass per unit length,

$\mu =\frac{m}{L}$

$$\begin{gathered} \mu =\frac{\left({\displaystyle \frac{\rho \pi L{D}^2}{4}}\right)}{L} \\ \mu =\left(\frac{\rho \pi D^2}{4}\right) \end{gathered}$$

Step 3. Finding the frequency, $f$

We know that Fundamental Frequency,

$f=\frac{1}{2L}\times \sqrt{\frac{T}{\mu }}$ [ $\mu =\frac{\rho {\mathrm{\pi D}}^2}{4}$]

$$\begin{gathered} f=\frac{1}{2L}\times \sqrt{\frac{T}{{\displaystyle \frac{\mathrm{\rho }\pi D^2}{4}}}} \\ f=\frac{1}{2L}\times \sqrt{\frac{4T}{{\mathrm{\rho \pi D}}^2}} \\ f=\frac{1}{2L}\times \frac{2}{D}\sqrt{\frac{T}{\mathrm{\rho \pi }}} \\ f=\frac{1}{LD}\sqrt{\frac{T}{\mathrm{\rho \pi }}} \\ \mathrm{f}\propto \frac{1}{\mathrm{LD}} \end{gathered}$$

Hence, option A is correct.