A source of sound is travelling with a velocity of 30 m/s towards a stationary observer. If actual frequency of source is 1000 Hz and the wind is blowing with velocity 20 m/s in a direction at $60^{\circ}$ with the direction of motion of source, then the apparent frequency heard by observer is (speed of sound is 340 m/s)

1011 Hz

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1000 Hz

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1094 Hz

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1086 Hz

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Solution

The correct option is C 1094 Hz
$f = (\frac{v + v_{m}}{v + v_{m} - v_{s o u r c e}}) 1000$
$= (\frac{340 + 20 c o s 60^{\circ}}{340 + 20 c o s 60^{\circ} - 30}) 1000$
=1094 Hz

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Similar questions

Q. A source of sound is travelling with a velocity of 30 m/s towards a stationary observer. If actual frequency of source is 1000 Hz and the wind is blowing with velocity 20 m/s in a direction at

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The correct option is C 1094 Hzf=(v+vmv+vm−vsource)1000 =(340+20cos60∘340+20 cos60∘−30)1000 =1094 Hz

The correct option is C 1094 Hz
$f = (\frac{v + v_{m}}{v + v_{m} - v_{s o u r c e}}) 1000$
$= (\frac{340 + 20 c o s 60^{\circ}}{340 + 20 c o s 60^{\circ} - 30}) 1000$
=1094 Hz