Question

A sound source moves along an xaxis, between detectors Aand B. The wavelength of the sound detected at Ais$\mathbf{0}\mathbf{.500}$that of the sound detected at B. What is the ratio${\mathbf{v}}_{\mathbf{s}}\mathbf{/}\mathbf{v}$of the speed of the source to the speed of sound?

Short answer

The ratio of the speed of source to the speed of sound will be $\frac{1}{3}$.

Step-by-step solution

The given data

The wavelength detected at detector A to that of detector B, ${\mathrm{\lambda }}_{\mathrm{B}}=0.5{\mathrm{\lambda }}_{\mathrm{A}}$.

Understanding the concept of the wave equations

We are given the wavelengths detected at A and B. Using these and the formula for frequency in terms of velocity, we can find the ratio of the speed of the source to the speed of sound.

Formula:

The frequency of a wave,

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

The frequency received by the observer according to Doppler’s Effect, (for observer at rest and source moving towards it)

${\mathbf{f}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{f}\left(\frac{v}{v-v_S}\right)$ …(ii)

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

For detector A, using equation (ii), we get

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

For detector B, using equation (ii), we get

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

We take the ratio for the above two equations of the frequencies, we get

$\frac{{\mathrm{f}}_{\mathrm{B}}^{\text{'}}}{{\mathrm{f}}_{\mathrm{A}}^{\text{'}}}=\frac{\mathrm{f}\left(\frac{\mathrm{v}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{S}}}\right)}{\mathrm{f}\left(\frac{\mathrm{v}}{\mathrm{v}-{\mathrm{v}}_{\mathrm{S}}}\right)}$

$\frac{{\mathrm{f}}_{\mathrm{B}}^{\text{'}}}{{\mathrm{f}}_{\mathrm{A}}^{\text{'}}}=\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{S}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{S}}}$ …(a)

Using value of frequencies from equation (i) in equation (a), we get

$\begin{array}{c}\frac{\left(\frac{\mathrm{v}}{{\mathrm{\lambda }}_{\mathrm{B}}}\right)}{\left(\frac{\mathrm{v}}{{\mathrm{\lambda }}_{\mathrm{A}}}\right)}=\frac{\frac{(1-{\mathrm{v}}_{\mathrm{S}})}{\mathrm{v}}}{\frac{(1+{\mathrm{v}}_{\mathrm{S}})}{\mathrm{v}}}\\ \frac{{\mathrm{\lambda }}_{\mathrm{B}}}{{\mathrm{\lambda }}_{\mathrm{A}}}=\frac{1-\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)}{1+\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)}\end{array}$

$\because {\mathrm{\lambda }}_{\mathrm{B}}=0.5{\mathrm{\lambda }}_{\mathrm{A}}$

So,

$\begin{array}{c}\frac{0.5{\mathrm{\lambda }}_{\mathrm{A}}}{{\mathrm{\lambda }}_{\mathrm{A}}}=\frac{1-\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)}{1+\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)}\\ 0.5=\frac{1-\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)}{1+\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)}\end{array}$

$\begin{array}{c}\frac{1}{2}=\frac{1-\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)}{1+\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)}\\ \frac{1}{2}\times \left(1+\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)\right)=1-\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)\\ \frac{1}{2}+\frac{1}{2}\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)=1-\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)\\ \frac{3}{2}\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)=1-\frac{1}{2}\\ \frac{3}{2}\left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)=\frac{1}{2}\\ \left(\frac{{\mathrm{v}}_{\mathrm{s}}}{\mathrm{v}}\right)=\frac{2}{3}\times \frac{1}{2}\\ =\frac{1}{3}\end{array}$

Hence, the value of the required ratio is 1/3.