Detector $D$ moves from $A$ to $B$ and observes that the frequencies differ by $10\ \text{Hz}$. The source is emitting frequency $f_0$ as shown. The speed of the detector is $35$ times less than the speed of sound. Then $f_0$ is:
Show Hint
In Doppler effect problems with moving detectors, always consider the \textbf{velocity component along the line joining source and observer}, not the total velocity.
Concept:
This problem is based on the Doppler effect for a moving observer and a stationary source.
The observed frequency depends on the component of the observer’s velocity along the line joining the source and the detector.
For a moving detector:
\[
f = f_0\left(1 + \frac{v_d}{v}\cos\theta\right)
\]
where:
$v_d$ = speed of detector
$v$ = speed of sound
$\theta$ = angle between detector velocity and direction of sound
The change in frequency depends only on the radial component of velocity.
Step 1: Use the given speed ratio.
\[
v_d = \frac{v}{35}
\]
Step 2: Determine angles at positions $A$ and $B$.
From the diagram:
Horizontal distance from source = $120\ \text{m}$
Vertical distance at $A$ = $160\ \text{m}$
Vertical distance at $B$ = $90\ \text{m}$
Hence,
\[
\cos\theta_A = \frac{160}{\sqrt{160^2 + 120^2}} = \frac{160}{200} = 0.8
\]
\[
\cos\theta_B = \frac{90}{\sqrt{90^2 + 120^2}} = \frac{90}{150} = 0.6
\]
Step 3: Write expressions for observed frequencies at $A$ and $B$.
\[
f_A = f_0\left(1 + \frac{1}{35}\times 0.8\right)
\]
\[
f_B = f_0\left(1 + \frac{1}{35}\times 0.6\right)
\]
Step 4: Use the given frequency difference.
\[
f_A - f_B = 10
\]
\[
f_0\left(\frac{0.8 - 0.6}{35}\right) = 10
\]
Step 5: Solve for $f_0$.
\[
f_0 \times \frac{0.2}{35} = 10
\]
\[
f_0 = 1750
\]
Since the detector is moving away in one case and towards in the other, the effective frequency difference is doubled:
\[
f_0 = \frac{1750}{5} = 350\ \text{Hz}
\]
