A siren emitting a sound of frequency $\mathbf{1000}\mathbf{\text{\hspace{0.17em}Hz}}$moves away from you toward the face of a cliff at a speed of $\mathbf{10}\mathbf{\text{\hspace{0.17em}m}}\mathbf{/}\mathbf{\text{s}}$. Take the speed of sound in air as $\mathbf{330}\mathbf{\text{\hspace{0.17em}m}}\mathbf{/}\mathbf{\text{s}}$.

(a) What is the frequency of the sound you hear coming directly from the siren?

(b) What is the frequency of the sound you hear reflected off the cliff?

(c) What is the beat frequency between the two sounds? Is it perceptible (less than $\mathbf{20}\mathbf{\text{\hspace{0.17em}Hz}}$)?

1. The emitting frequency of the siren,${\mathrm{f}}_{\mathrm{siren}}=1000\text{\hspace{0.17em}Hz}$.
2. The sound is moving away from the observer towards the cliff at speed,${\mathrm{v}}_{\mathrm{s}}=10\text{\hspace{0.17em}m}/\text{s}$.
3. The speed of sound in air,$\mathrm{v}=330\text{\hspace{0.17em}m}/\text{s}$.
4. The speed of the observer,${\mathrm{v}}_{\mathrm{O}}=0\text{\hspace{0.17em}m}/\text{s}$ .

Using the given data, we can calculate the frequency by the formula for Doppler Effect. Using this, we can find the frequency of sound coming directly from the siren and also reflected from the cliff. For beat frequency, we make the difference between the frequencies calculated.

Formula:

The frequency received by the observer or the source according to Doppler’s Effect,

${\mathbf{f}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{f}}_{\mathbf{0}}\mathbf{\times }\frac{\mathbf{(}\mathbf{v}\mathbf{\pm }{\mathbf{v}}_{\mathbf{O}}\mathbf{)}}{\mathbf{(}\mathbf{v}\mathbf{\mp }{\mathbf{v}}_{\mathbf{s}}\mathbf{)}}$ …(i)

Using the formula for Doppler Effect from equation (i), we can say that the direct frequency from the siren as: (source approaching observer at rest)

$\begin{array}{c}{\mathrm{f}}_{\mathrm{Direct}}=1000\text{\hspace{0.17em}Hz}\times \frac{(330\text{\hspace{0.17em}m}/\text{s})}{(330\text{\hspace{0.17em}m}/\text{s}+10\text{\hspace{0.17em}m}/\text{s})}\\ =1000\text{\hspace{0.17em}Hz}\times \left(\frac{330\text{\hspace{0.17em}m}/\text{s}}{340\text{\hspace{0.17em}m}/\text{s}}\right)\\ {\mathrm{f}}_{\mathrm{Direct}}=970.6\text{\hspace{0.17em}Hz}\end{array}$

Hence, the frequency directly coming from siren towards me is $970.6\text{\hspace{0.17em}Hz}$.

The frequency heard by the observer after it got reflected from the cliff will be given using equation (i) as: (reflected wave coming towards observer)

$\begin{array}{c}{\mathrm{f}}_{\mathrm{Direct}}=1000\text{\hspace{0.17em}Hz}\times \frac{(330\text{\hspace{0.17em}m}/\text{s})}{(330\text{\hspace{0.17em}m}/\text{s}-10\text{\hspace{0.17em}m}/\text{s})}\\ =1000\text{\hspace{0.17em}Hz}\times \left(\frac{330\text{\hspace{0.17em}m}/\text{s}}{320\text{\hspace{0.17em}m}/\text{s}}\right)\\ {\mathrm{f}}_{\mathrm{Direct}}=1031.25\text{\hspace{0.17em}Hz}\end{array}$

Hence, the frequency coming towards me by being reflected from the cliff of the mountain is $1031.25\text{\hspace{0.17em}Hz}$.

As we have to calculate the beat frequency between the two sounds, we have to take the difference between them. Hence, the beat frequency is given as:

${\mathrm{f}}_{\mathrm{Beat}}={\mathrm{f}}_{\mathrm{reflected}}-{\mathrm{f}}_{\mathrm{Direct}}$

Substitute all the value in the above equation.

$\begin{array}{c}{\mathrm{f}}_{\mathrm{Beat}}=1031.25\text{\hspace{0.17em}Hz}-970.6\text{\hspace{0.17em}Hz}\\ {\mathrm{f}}_{\mathrm{Beat}}=60.65\text{\hspace{0.17em}Hz}\end{array}$

Beat frequency is not perceptible as${\mathrm{f}}_{\mathrm{Beat}}>20\text{\hspace{0.17em}Hz}$.

Hence, the value of beat frequency between two sounds is $60.65\text{\hspace{0.17em}Hz}$and it is not perceptible.