Question

"An airplane traveling at one third the speed of sound (i.e., 114 m/s) emits a sound of frequency 3.65 kHz. At what frequency does a stationary listener hear the sound as the plane approaches? Answer in units of kHz."

Answer 1

Answer:

Explanation:

velocity of source, Vs = 114 m/s

original frequency, f = 3.65 kHz

velocity of observer , Vo = 0 m/s

velocity of sound, v = 334 m/s

let the frequency is f'.

The formula for the doppler effect is given by

$f'=\left ( \frac{v+v_{0}}{v-v_{s}} \right )f$

where, v is the velocity of sound.

$f'=\left ( \frac{334+0}{334-114} \right )\times 3.65$

f' = 5.54 kHz

Answer 2

The frequency at which the stationary listener will hear the sound of the plane as it approaches is 5.54 KHz

Data obtained from the question

• velocity of source (vₛ) = 114 m/s
• original frequency (f₀) = 3.65 KHz
• velocity of observer (v₀) = 0 m/s
• velocity of sound (v) = 334 m/s

• Observer's frequency (f₀) =?

How to determine the the observer's frequency

Applying the equation of Doppler effect, the observer's frequency can be obtained as follow:

f₀ = [(v + v₀) / (v – vₛ)]f

f₀ = [(334 + 0) / (334 – 114)] × 3.65

f₀ = [334 / 220] × 3.65

f₀ = 5.54 KHz

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