Question

A steel wire of mass $4.0g$ and length $80cm$ is fixed at the two ends. The tension in the wire is $50N$. Find the frequency and wavelength of the fourth harmonic of the fundamental.

Answer 1

Step 1 Given data.

Mass of a steel wire, $m=4.0g=0.004kg$

Length of the wire, $L=80cm=0.8m$

Tension in the wire, $T=50N$

Step 2: Finding the frequency.

We know that,

$f=\frac{v}{2L}$

Where, $f$ is fundamental frequency, $v$ is the speed of the wave.

And,

$f_0=\frac{\sqrt{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.}}}{2L}$ $\left[\because v=\sqrt{\frac{T}{\rho }}\right]$

Where, $T$ is the tension and $\rho$ is mass per unit length.

$f_0=\frac{\sqrt{{\displaystyle \raisebox{1ex}{$50$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \raisebox{1ex}{$0.004$}\!\left/ \!\raisebox{-1ex}{$0.8$}\right.}$}\right.}}}{2\times 0.8}$ $\left[\because \rho =\frac{mass}{length}\right]$

$\Rightarrow$$f_0=62.5Hz$

Step 3: Finding the frequency of the fourth harmonic of the fundamental.

$f_n=n\times f_0$

Where $f_n$ is the $nth$harmonic frequency, $n$ is the number of harmonic, $f_0$ is the fundamental frequency.

At $n=4$,

Then, the frequency of 4th harmonic.

$f_4=4\times 62.5$

$\Rightarrow$$f_4=250Hz$

Step 4: Finding the wavelength of the fourth harmonic of the fundamental.

As we know,

$v_4=f_4{\lambda }_4$

Where ${\lambda }_4$ fourth harmonic wavelength,$v_4$ is speed of the fourth harmonic wave.

$\sqrt{\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$\rho $}\right.}=f_4{\lambda }_4$ $\left[\because v=\sqrt{\frac{T}{\rho }}\right]$

$\Rightarrow$ ${\lambda }_4=\frac{\sqrt{\raisebox{1ex}{$50$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \raisebox{1ex}{$0.004$}\!\left/ \!\raisebox{-1ex}{$0.8$}\right.}$}\right.}}{250}$ $\left[\because f_4=250Hz\right]$

$\Rightarrow$${\lambda }_4=0.4m$

Hence, the frequency of the fourth harmonic is $250Hz$ and the wavelength of the fourth harmonic of the fundamental is $0.4m$.