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Two engines pass each other moving in op...

Two engines pass each other moving in opposite directions with uniform speed of 30m/so .one of them is blowing a whistle of frequency 540 Hz. Calculate the frequency heard by driver of second engine before they pass each other. Speed of sound is 300 m/sec :

A

648 Hz

B

450 Hz

C

270 Hz

D

540 Hz

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The correct Answer is:
To solve the problem of finding the frequency heard by the driver of the second engine before they pass each other, we will use the Doppler effect formula for sound. Here’s a step-by-step solution: ### Step 1: Identify the known values - Frequency of the whistle (f₀) = 540 Hz - Speed of sound (v) = 300 m/s - Speed of the observer (v₀) = 30 m/s (the driver of the second engine) - Speed of the source (vₛ) = 30 m/s (the engine blowing the whistle) ### Step 2: Apply the Doppler effect formula The formula for the frequency heard by the observer when the source and observer are moving towards each other is given by: \[ f' = f_0 \left( \frac{v + v_0}{v - v_s} \right) \] Where: - \( f' \) = frequency heard by the observer - \( f_0 \) = frequency of the source (540 Hz) - \( v \) = speed of sound (300 m/s) - \( v_0 \) = speed of the observer (30 m/s) - \( v_s \) = speed of the source (30 m/s) ### Step 3: Substitute the known values into the formula Substituting the known values into the formula: \[ f' = 540 \left( \frac{300 + 30}{300 - 30} \right) \] ### Step 4: Simplify the expression Calculate the numerator and denominator: - Numerator: \( 300 + 30 = 330 \) - Denominator: \( 300 - 30 = 270 \) Now substituting these values back into the equation: \[ f' = 540 \left( \frac{330}{270} \right) \] ### Step 5: Simplify the fraction Now simplify \( \frac{330}{270} \): \[ \frac{330}{270} = \frac{11}{9} \] ### Step 6: Calculate the final frequency Now substituting back into the equation: \[ f' = 540 \times \frac{11}{9} \] Calculating this gives: \[ f' = 540 \times 1.2222 \approx 660 \text{ Hz} \] ### Conclusion The frequency heard by the driver of the second engine before they pass each other is **660 Hz**. ---
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