Question

Two sound waves with amplitude of$\mathbf{12}\mathbf{\text{\hspace{0.17em}nm}}$and a wavelength of$\mathbf{35}\mathbf{\text{\hspace{0.17em}cm}}$travel in the same direction through a long tube, with a phase difference of$\mathit{\pi }\mathbf{/}\mathbf{3}\mathbf{\text{\hspace{0.17em}rad}}$. What are the (a) amplitude and (b) wavelength of the net sound wave produced by their interference? If, instead, the sound waves travel through the tube in opposite directions, what are the (c) amplitude and (d) wavelength of the net wave?

Short answer

1. Amplitude of the net sound wave produced by their interference will be $21\text{\hspace{0.17em}nm}$.
2. Wavelength of the net sound wave produced by their interference will be $35\text{\hspace{0.17em}cm}$.
3. Amplitude of the net wave when the waves travel through the pipe in opposite direction will be $24\text{\hspace{0.17em}nm}$.
4. Wavelength of the net wave when the waves travel through the pipe in opposite direction will be $35\text{\hspace{0.17em}cm}$.

Step-by-step solution

The given data

1. Amplitude of the two waves, $y_m=12\text{\hspace{0.17em}nm}$.
2. Wavelength of the two waves,$\lambda =35\text{\hspace{0.17em}cm}$.
3. Phase difference between them, $\varphi =\pi /3$.

Understanding the concept of the wave

The amplitude and the phase difference for the two waves are given. Using this we can find the resultant wave. Once we get the resultant wave, we can find the required quantities.

Formula:

The expression of the displacement of the wave,

$\mathit{y}\mathbf{=}{\mathit{y}}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{k}\mathbf{x}\mathbf{-}\mathbf{\omega }\mathbf{t}\mathbf{)}$ …(i)

The resultant wave equation,

$\mathit{Y}\mathbf{=}\mathbf{2}{\mathit{y}}_{\mathbf{m}}\mathit{c}\mathit{o}\mathit{s}\left(\frac{{\displaystyle \varphi }}{{\displaystyle 2}}\right)\mathit{s}\mathit{i}\mathit{n}\left(kx-\omega t+\frac{{\displaystyle \varphi }}{{\displaystyle 2}}\right)$ …(ii)

a) Calculation of the amplitude of net sound wave when waves travel in same direction

The amplitude and the phase difference between the two waves is given, so, using equation (i), we can write the displacement equations as:

$y_1=y_m\sin (kx-\omega t)$

$y_2=y_m\sin (kx-\omega t+\varphi )$

From equation (ii), the resultant amplitude is given as:

$\begin{array}{c}{Y}_m=2\times (12\text{\hspace{0.17em}nm})\times \cos \left(\frac{\pi }{6}\right)\\ =20.78\text{\hspace{0.17em}nm}\\ Y_m\approx 21\text{\hspace{0.17em}nm}\end{array}$

Hence, the amplitude of the resultant amplitude during interference is $21\text{\hspace{0.17em}nm}$.

b) Calculation of the wavelength of net sound wave when waves travel in same direction 

Wavelength of the superimposed waves do not change, hence, the resultant wavelength would also bethesame asthewavelength of the sound waves.

${\lambda }_{resul\tan t}=35\text{\hspace{0.17em}cm}$

Hence, the value of the resultant wavelength is $35\text{\hspace{0.17em}cm}$.

c) Calculation of the amplitude of net sound wave when waves travel in opposite direction

As the waves are travelling in the opposite direction

The phase difference between them will be$\varphi =\pi$

Using equation (ii) and the given values, we get the amplitude as:

$Y_m=2y_m\cos \left(\varphi \right)$

Substitute all the value in the above equation.

$\begin{array}{l}{Y}_m=2\times 12\text{\hspace{0.17em}nm}\times \cos \left(\frac{\pi }{2}\right)\\ =|2\times 12\text{\hspace{0.17em}nm}\times \left(1\right)|\\ Y_m=24\text{\hspace{0.17em}nm}\end{array}$

Hence, the value of net amplitude is$24\text{\hspace{0.17em}nm}$ .

d) Calculation of the wavelength of net sound wave when waves travel in opposite direction

Change in direction would have no effect on the wavelength of the resultant wave

${\lambda }_{resul\tan t}=35\text{\hspace{0.17em}cm}$

Hence, the value of the wavelength is $35\text{\hspace{0.17em}cm}$.