# A person who is standing feels 2% difference of frequency of a car horn. If a car is moving to the person and the velocity of the sound is 332 m/s, th

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A person who is standing feels 2% difference of frequency of a car horn. If a car is moving to the person and the velocity of the sound is 332 m/s, then the velocity of car will be:
1. 6.5 m/s
2. 8 m/s
3. 4 m/s
4. 9.2 m/s

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Correct Answer - Option 1 : 6.5 m/s

CONCEPT:

Doppler effect:

• Doppler effect in physics is defined as the increase (or decrease) in the frequency of sound, light, or other waves as the source and the observer move towards (or away from) each other.
• Waves emitted by a source traveling towards an observer gets compressed. In contrast, waves emitted by a source traveling away from an observer get stretched out.
• Doppler Effect (Doppler Shift) was first proposed by Christian Johann Doppler in 1842.
• Doppler effect formula: When the source and the observer moving towards each other,

$⇒ f'=\frac{(v+ v_{o})}{(v- v_{s})}× f$

Where f' = apparent frequency(Hz), f = actual frequency(Hz), v = velocity of the sound wave(m/s), vo = velocity of the observer(m/s), and vs = velocity of the sound(m/s)

CALCULATION:

Given %Δf = 2%, vo = 0 m/s and v = 332 m/s

• The apparent frequency is given as,

$⇒ f'=\frac{(v+ v_{o})}{(v- v_{s})}× f$

$⇒ f'=\frac{(332+ 0)}{(332- v_{s})}× f$

$⇒ f'=\frac{332}{(332- v_{s})}× f$     -----(1)

• The percentage change in frequency is given as,

⇒ %Δf = 2%

$⇒ \frac{f'-f}{f}\times100=2$     -----(2)

By equation 1, and equation 2,

$⇒ \frac{\frac{332}{(332- v_{s})}× f-f}{f}\times100=2$

$⇒ \left ( \frac{332}{332-v_{s}}-1 \right )\times100=2$

$⇒ \frac{v_{s}}{332-v_{s}}\times100=2$

⇒ 100vs = 664 - 2vs

⇒ 102vs = 664

⇒ vs = 6.5 m/s

• Hence, option 1 is correct.