Question

A sinusoidal sound wave moves at$\mathbf{343}\mathbf{\text{\hspace{0.17em}m}}\mathbf{/}\mathbf{\text{s}}$through air in the positive direction of an xaxis. At one instant during the oscillations, air molecule Ais at its maximum displacement in the negative direction of the axis while air molecule Bis at its equilibrium position. The separation between those molecules is$\mathbf{15}\mathbf{.0}\mathbf{\text{\hspace{0.17em}cm}}$, and the molecules between Aand Bhave intermediate displacements in the negative direction of the axis. (a) What is the frequency of the sound wave?

In a similar arrangement but for a different sinusoidal sound wave, at one instant air molecule Cis at its maximum displacement in the positive direction while molecule Dis at its maximum displacement in the negative direction. The separation between the molecules is again$\mathbf{15}\mathbf{.0}\mathbf{\text{\hspace{0.17em}cm}}$, and the molecules between Cand Dhave intermediate displacements. (b) What is the frequency of the sound wave?

Short answer

1. Frequency of the sound wave for air molecules A and B is $572\text{\hspace{0.17em}Hz}$.
2. Frequency of the sound wave for air molecules C and D is $1144\text{\hspace{0.17em}Hz}$.

Step-by-step solution

The given data

1. Distance between molecule A and B is $d=15.0\text{\hspace{0.17em}cmor}0.15\text{\hspace{0.17em}m}$.
2. Distance between molecule C and D is$d=15.0\text{\hspace{0.17em}cmor}0.15\text{\hspace{0.17em}m}$.
3. Velocity of sound, $v=343\text{\hspace{0.17em}m}/\text{s}$.

Understanding the concept of the frequency

Using the fact that the positions of the molecule A and B are given, we can find the wavelength. From the velocity for sound and wavelength, we can find the frequency of the wave using the corresponding relation. The same procedure can be used for molecules C and D.

Formula:

The frequency of the oscillation wave,

$\mathit{f}\mathbf{=}\mathit{v}\mathbf{/}\mathit{\lambda }$ …(i)

a) Calculation of the frequency of sound wave for molecules A and B

The molecule A is at maximum displacement in the negative direction of the axis and molecule B is at the equilibrium position. Hence, the wavelength is given as:

$\begin{array}{c}d=\frac{\lambda }{4}\\ \lambda =4d\end{array}$

Substitute all the value in the above equation.

role="math" localid="1661574315309" $\begin{array}{l}\lambda =4\times 0.15\text{\hspace{0.17em}m}\\ \lambda =0.6\text{\hspace{0.17em}m}\end{array}$

Using equation (i), the frequency of the wave for molecules A and B is given as:

$\begin{array}{l}f=\frac{343\text{\hspace{0.17em}m}/\text{s}}{0.6\text{\hspace{0.17em}m}}\\ =571.67\text{\hspace{0.17em}Hz}\\ ~572\text{\hspace{0.17em}Hz}\end{array}$

Hence, the value of the frequency of the sound wave is $572\text{\hspace{0.17em}Hz}$.

b) Calculation of the frequency of sound wave for molecules C and D

The molecule C is at maximum displacement in the positive direction of the axis, and Molecule D is at maximum displacement in the negative direction of the axis. From this we can write, the value of wavelength as:

$\begin{array}{c}d=\frac{\lambda }{2}\\ \lambda =2d\end{array}$

Substitute all the value in the above equation.

role="math" localid="1661574883069" $\begin{array}{l}\lambda =2\times 0.15\text{\hspace{0.17em}m}\\ \lambda =0.3\text{\hspace{0.17em}m}\end{array}$

Using equation (i), the frequency of the wave for molecules C and D is given as:

$\begin{array}{c}f=\frac{343\text{\hspace{0.17em}m}/\text{s}}{0.3\text{\hspace{0.17em}m}}\\ =1143.33\text{\hspace{0.17em}Hz}\\ ~1144\text{\hspace{0.17em}Hz}\end{array}$

Hence, the value of the frequency of the sound wave is $1144\text{\hspace{0.17em}Hz}$.