Question

An avalanche of sand along some rare desert sand dunes can produce a booming that is loud enough to be heard $\mathbf{10}\mathbf{\text{\hspace{0.17em}km}}$away. The booming apparently results from a periodic oscillation of the sliding layer of sand — the layer’s thickness expands and contracts. If the emitted frequency isrole="math" localid="1661512432408" $\mathbf{90}\mathbf{\text{\hspace{0.17em}Hz}}$, what are (a) the period of the thickness oscillation and (b) the wavelength of the sound?

Short answer

1. The period of the thickness oscillation is $1.1\times {10}^{-2}\text{\hspace{0.17em}s}$.
2. The wavelength of the sound is $3.8\text{\hspace{0.17em}m}$.

Step-by-step solution

The given data

The emitted frequency,$\mathrm{f}=90\text{\hspace{0.17em}Hz}$.

The sand dunes can produce a booming upto distance,$\mathrm{d}=10\text{\hspace{0.17em}kmor}10000\text{\hspace{0.17em}m}$ .

Understanding the concept of the wave equations

Using the relation between period and frequency, we can find the period of thickness oscillation. Using the formula of wave speed, we can find the wavelength from the given frequency of the wave.

Formula:

The period of oscillations of a wave,

$\mathbf{T}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{f}$ …(i)

The velocity of a wave,

$\mathbf{v}\mathbf{=}\mathbf{\lambda f}$ …(ii)

a) Calculation of the period of the thickness oscillation

We have relation between frequency and time period from equation (i), hence, the period of oscillation is given as:

$\begin{array}{c}\mathrm{T}=\frac{1}{90\text{\hspace{0.17em}Hz}}\\ \mathrm{T}=1.1\times {10}^{-2}\text{\hspace{0.17em}s}\end{array}$

Therefore, the period of the thickness oscillation is $1.1\times {10}^{-2}\text{\hspace{0.17em}s}$.

b) Calculation of wavelength of sound

The velocity of sound is,$\mathrm{v}=343\text{\hspace{0.17em}m}/\text{s}$.

Using equation (ii), we can get the wavelength of a sound wave as:

$\begin{array}{c}\mathrm{\lambda }=\frac{\mathrm{v}}{\mathrm{f}}\\ =\frac{343\text{\hspace{0.17em}m}/\text{s}}{90\text{\hspace{0.17em}Hz}}\\ \mathrm{\lambda }=3.8\text{\hspace{0.17em}m}\end{array}$

Therefore, the wavelength of the sound is $3.8\text{\hspace{0.17em}m}$.