Question

A detector initially moves at constant velocity directly toward a stationary sound source and then (after passing it) directly from it. The emitted frequency is $\mathit{f}$. During the approach the detected frequency is${\mathit{f}}_{\mathbf{a}\mathbf{p}\mathbf{p}}^{\mathbf{\text{'}}}$ and during the recession it is${\mathbf{f}}_{\mathbf{rec}}^{\mathbf{\text{'}}}$ . If the frequencies are related by $\mathbf{(}{\mathbf{f}}_{\mathbf{app}}^{\mathbf{\text{'}}}\mathbf{-}{\mathbf{f}}_{\mathbf{rec}}^{\mathbf{\text{'}}}\mathbf{)}\mathbf{/}\mathbf{f}\mathbf{}\mathbf{=}\mathbf{0.500}$, what is the ratio ${\mathbf{v}}_{\mathbf{D}}\mathbf{/}\mathbf{v}$of the speed of the detector to the speed of sound?

Short answer

The ratio ${\mathrm{v}}_{\mathrm{D}}/\mathrm{v}$ of the speed of the detector to the speed of sound is $0.250$.

Step-by-step solution

The given data

The frequencies are related by,

$\frac{{\mathrm{f}}_{\mathrm{app}}^{\text{'}}-{\mathrm{f}}_{\mathrm{rec}}^{\text{'}}}{\mathrm{f}}=0.500$

Understanding the concept of the Doppler Effect

We use Doppler Effect to write the equation when the detector approaches the source and when the detector is moving away from the source. Using these equations, we can find the ratio ${\mathbf{v}}_{\mathbf{D}}\mathbf{/}\mathbf{v}$of the speed of the detector to the speed of sound.

Formula:

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

${\mathbf{f}}^{\mathbf{\text{'}}}\mathbf{=}\left(\frac{V\pm V_l}{V\pm V_s}\right)\mathbf{}\mathbf{f}$ …(i)

Calculation of the ratio of speed of the detector to that of the speed of sound

Forthe detector which approaches the source, we have the frequency using equation (i) as:

As ${\mathrm{v}}_{\mathrm{s}}=0\text{\hspace{0.17em}m}/\text{s}$

${\mathrm{f}}_{\mathrm{app}}^{\text{'}}=\mathrm{f}\left(\frac{\mathrm{v}+{\mathrm{v}}_{\mathrm{D}}}{\mathrm{v}}\right)$

And, when the detector is moving away from the source, the frequency using equation (i) is given as:

${\mathrm{f}}_{\mathrm{rec}}^{\text{'}}=\mathrm{f}\left(\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{D}}}{\mathrm{v}}\right)$

So, the difference of both frequencies is given as:

$\begin{array}{c}{\mathrm{f}}_{\mathrm{app}}^{\text{'}}-{\mathrm{f}}_{\mathrm{rec}}^{\text{'}}=\mathrm{f}\left(\frac{\mathrm{v}+{\mathrm{v}}_{\mathrm{D}}}{\mathrm{v}}\right)-\mathrm{f}\left(\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{D}}}{\mathrm{v}}\right)\\ \frac{({\mathrm{f}}_{\mathrm{app}}^{\text{'}}-{\mathrm{f}}_{\mathrm{rec}}^{\text{'}})}{\mathrm{f}}=\frac{\mathrm{v}}{\mathrm{v}}+\frac{{\mathrm{v}}_{\mathrm{D}}}{\mathrm{v}}-\frac{\mathrm{v}}{\mathrm{v}}+\frac{{\mathrm{v}}_{\mathrm{D}}}{\mathrm{v}}\\ \frac{({\mathrm{f}}_{\mathrm{app}}^{\text{'}}-{\mathrm{f}}_{\mathrm{rec}}^{\text{'}})}{\mathrm{f}}=2\frac{{\mathrm{v}}_{\mathrm{D}}}{\mathrm{v}}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\frac{1}{2}=2\frac{{\mathrm{v}}_{\mathrm{D}}}{\mathrm{v}}\\ \frac{{\mathrm{v}}_{\mathrm{D}}}{\mathrm{v}}=\frac{1}{4}\\ =0.25\end{array}$

Therefore, the ratio ${\mathrm{v}}_{\mathrm{D}}/\mathrm{v}$ of the speed of the detector to the speed of sound is $0.250$.