91Ó°ÊÓ

• i)The speed of the source is$V_S=15\frac{m}{s}$
• ii)The speed of the listener is$V_D=25\frac{m}{S}$
• iii) The difference in frequency is$f_2-f_1=37Hz$

Use the principle of the Doppler Effect and its corresponding equations to get the equations of ${\mathrm{f}}_1\mathrm{and}{\mathrm{f}}_2$in terms of f. Then we can use the value of the difference of${\mathrm{f}}_1\mathrm{and}{\mathrm{f}}_2$and solve for the frequency of source f.

Formula:

The frequency received by the observer or the source according to Doppler’s Effect, (Consider detector at rest and source is approaching detector)

${\mathrm{f}}_1=\mathrm{f}\frac{\mathrm{V}}{\mathrm{V}-{\mathrm{V}}_{\mathrm{S}}}$ (i)

(Consider source is approaching detector and detector is approaching source)

${\mathrm{f}}_2=\mathrm{f}×\frac{\mathrm{V}+{\mathrm{V}}_{\mathrm{D}}}{\mathrm{V}-{\mathrm{V}}_{\mathrm{S}}}\left(\mathrm{ii}\right)$

Consider the equation (i) of Doppler Effect for stationary detector and moving source as follows:

${\mathrm{f}}_1=\mathrm{f}\frac{340}{340-15}$ ….. (1)

Now, the source as well asthedetector is moving towards each other. So, using equation (ii), we get

${\mathrm{f}}_2=\mathrm{f}×\frac{340+25}{340-15}$ ….. (2)

Now, we have the frequency difference as:

$$\begin{gathered} \mathrm{f}×\frac{340+25}{340-15}-\mathrm{f}×\frac{340}{340-15}=37\mathrm{Hz} \\ \mathrm{f}\left(\frac{340+25}{340-15}-\frac{340}{340-15}\right)=37\mathrm{Hz} \\ \mathrm{f}\left(\frac{25}{325}\right)=37\mathrm{Hz} \\ \mathrm{f}=481\mathrm{Hz} \end{gathered}$$

Hence, the value of the frequency is 481Hz.