Two cars moving in opposite directions approach each other with speed of $22 ext{ m/s}$ and $16.5 ext{ m/s

Question

Two cars moving in opposite directions approach each other with speed of $22 	ext{ m/s}$ and $16.5 	ext{ m/s}$ respectively. The driver of the first car blows a horn having a frequency $400 	ext{ Hz}$. The frequency heard by the driver of the second car is [velocity of sound $340 	ext{ m/s}$]

Accepted Answer

1. **Identify the Doppler Effect Formula:** The apparent frequency ($f'$) heard by an observer when both the source and observer are in relative motion along the line joining them is given by $f' = f \left( \frac{v \pm v_o}{v \mp v_s} ight)$, where $v$ is the velocity of sound in the medium, $v_o$ is the velocity of the observer, and $v_s$ is the velocity of the source .
2. **Determine Signs for Approaching Bodies:** Since both cars are moving towards each other, they both contribute to an increase in the apparent frequency. 
   - The observer approaching the source means the numerator is $(v + v_o)$.
   - The source approaching the observer means the denominator is $(v - v_s)$.
3. **Substitute the Given Values:**
   Given $v = 340 	ext{ m/s}$, $v_o = 16.5 	ext{ m/s}$ (second car), $v_s = 22 	ext{ m/s}$ (first car), and $f = 400 	ext{ Hz}$.
   $$f' = 400 \left( \frac{340 + 16.5}{340 - 22} ight)$$
   $$f' = 400 \left( \frac{356.5}{318} ight)$$
   $$f' = 400 	imes 1.12107... \approx 448.4 	ext{ Hz}$$
Rounding off to the nearest integer, the frequency heard by the driver of the second car is $448 	ext{ Hz}$.

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