Doppler Effect in Sound with Wind Velocity for Approaching Vehicles

To determine the relation between the observed frequency $f_2$ and the emitted frequency $f_1$, we apply the Doppler Effect formula for sound, taking into account the effect of wind.

1. General Formula for Doppler Effect with Wind Velocity

Let:

• $V$ = speed of sound in still air
• $w$ = speed of wind parallel to the road
• $u$ = speed of both vehicles (source $S$ and observer $O$), moving towards each other
• $v_{\text{sound}}$ = effective velocity of sound from the source to the observer relative to the ground

The general Doppler Effect equation for the observed frequency $f_2$ when both source and observer are in motion is given by: $f_2 = f_1 \left( \frac{v_{\text{sound}} + v_O}{v_{\text{sound}} - v_S} \right)$ where:

• $v_O$ is the component of the velocity of the observer in the direction towards the source.
• $v_S$ is the component of the velocity of the source in the direction towards the observer.

Since both vehicles are approaching each other, we have: $v_O = u \quad \text{and} \quad v_S = u$

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2. Case 1: Wind blows from Observer to Source

• The sound wave travels from the source $S$ to the observer $O$.
• The wind is blowing in the opposite direction of sound propagation (from $O$ to $S$).
• Therefore, the effective speed of sound relative to the ground is: $v_{\text{sound}} = V - w$

Substituting $v_{\text{sound}}$, $v_O = u$, and $v_S = u$ into the Doppler formula: $f_2 = f_1 \left[ \frac{(V - w) + u}{(V - w) - u} \right]$

Since $u > 0$ and assuming the sound reaches the observer ($V - w > u$), the numerator is strictly greater than the denominator: $(V - w) + u > (V - w) - u$

Therefore: $\frac{(V - w) + u}{(V - w) - u} > 1 \implies f_2 > f_1$

Thus, Option (A) is correct.

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3. Case 2: Wind blows from Source to Observer

• The sound wave travels from the source $S$ to the observer $O$.
• The wind is blowing in the same direction as sound propagation (from $S$ to $O$).
• Therefore, the effective speed of sound relative to the ground is: $v_{\text{sound}} = V + w$

Substituting $v_{\text{sound}}$, $v_O = u$, and $v_S = u$ into the Doppler formula: $f_2 = f_1 \left[ \frac{(V + w) + u}{(V + w) - u} \right]$

Since $u > 0$, the numerator is again strictly greater than the denominator: $(V + w) + u > (V + w) - u$

Therefore: $\frac{(V + w) + u}{(V + w) - u} > 1 \implies f_2 > f_1$

Thus, Option (B) is correct.

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Conclusion

In both cases, because the source and observer are approaching each other, the observed frequency $f_2$ is greater than the source frequency $f_1$ irrespective of the direction of the wind.

Correct Options: A, B