Physics Waves PMDC 2020
PMDC Verified Question 96 of 125
A whistler with velocity 33 ms\(^{-1}\) approaches towards a stationary observer with frequency 450Hz what is the apparent frequency heard by the observer:
A
500
B
400
C
430
D
450
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: 500
Concept:

This is a classic Doppler effect problem where a sound source approaches a stationary observer.

Formula:

$$ f' = \left( \frac{v}{v - v_s} \right) f $$

Solution:

  • Assume standard velocity of sound in air \( v \approx 330 \text{ m/s} \).


  • Source velocity \( v_s = 33 \text{ m/s} \) and true frequency \( f = 450 \text{ Hz} \).


  • Substitute into the formula: \( f' = \left( \frac{330}{330 - 33} \right) 450 \).


  • Calculate the denominator: \( 330 - 33 = 297 \).


  • Simplify the fraction: \( \frac{330}{297} = \frac{10}{9} \).


  • Multiply by the base frequency: \( f' = \frac{10}{9} \times 450 = 10 \times 50 = 500 \text{ Hz} \).


Why other options are incorrect:

Option B would be correct if the source were receding (moving away). Option D is the original frequency, incorrectly assuming no Doppler shift occurs.

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