Question

A string under tension ${\mathit{\tau }}_{\mathbf{i}}$ oscillates in the third harmonic at frequency ${\mathit{f}}_{\mathbf{3}}$, and the waves on the string have wavelength ${\mathit{\lambda }}_{\mathbf{3}}$. If the tension is increased to ${\mathit{\tau }}_{\mathbf{f}}\mathbf{=}\mathbf{4}{\mathit{\tau }}_{\mathbf{i}}$ and the string is again made to oscillate in the third harmonic. What then are (a) the frequency of oscillation in terms of ${\mathit{f}}_{\mathbf{3}}$ and (b) the wavelength of the waves in terms of ${\mathit{\lambda }}_{\mathbf{3}}$?

Short answer

1. The frequency in oscillation is $2f_3$
2. The wavelength of the waves is ${\lambda }_3$

Step-by-step solution

Given data

The tension is increased to ${\tau }_f=4{\tau }_i$.

Understanding the concept of resonant frequency

By using the formulas for resonant frequency fand the speed v of the wave, we can find the frequency of oscillation in terms of and by using the formula for the new wavelength we can find the wavelength of the waves in terms of .

Formula:

The resonant frequency of a wave,$f=\frac{v}{\lambda }......\left(1\right)$

The speed of wave, $v=\sqrt{\frac{\tau }{\mu }}.........\left(2\right)$

The nth frequency of a wave, $f_n=\frac{nv}{2L}.....\left(3\right)$

Step 3(a): Calculation for the frequency in oscillation

Substituting the equation (2) in equation (3), we get the nth frequency as:

$f_n=\frac{n}{2L}\times \sqrt{\frac{\tau }{\mu }}$

Hence, the third frequency is given as:

$f_3=\frac{3}{2L}\times \sqrt{\frac{{\tau }_i}{\mu }}......\left(4\right)$

When,${\tau }_f=4{\tau }_i$, using this in equation (4),we get the new frequency is given as:

$$\begin{gathered} f{\text{'}}_3=\frac{3}{2L}\times \sqrt{\frac{{\tau }_i}{\mu }} \\ =2\left[\frac{3}{2L}\times \sqrt{\frac{{\tau }_i}{\mu }}\right](\because {\tau }_f=4{\tau }_i) \\ =2f_3 \end{gathered}$$

Hence, the value of the new frequency is $2f_3$

Step 4(b): Calculation for the wavelength

Using equation (1), we can get the new wavelength as given:

$$\begin{gathered} \lambda {\text{'}}_3=\frac{v\text{'}}{f{\text{'}}_3} \\ =\frac{\sqrt{{\displaystyle \frac{{\tau }_i}{\mu }}}}{2f_3}(\because u\sin gequation(2\left)\right) \\ =2\left[\begin{array}{c}\frac{\sqrt{{\displaystyle \frac{{\tau }_i}{\mu }}}}{2f_3}\\ \end{array}\right](\because f_3\text{'}=f_{3}and{\tau }_f=4{\tau }_i) \\ =\frac{v}{f_3} \\ ={\lambda }_3 \end{gathered}$$

Hence, the value of new wavelength is ${\lambda }_3$