# At what frequency does a stationary listener hear the sound 1) as the plane approaches and 2) after it passes?I'm stumped as to how to solve this. Please help me out. Also, please DO NOT ROUND...

At what **frequency** does a stationary listener hear the sound 1) ** as the plane approaches** and 2)

**?**

*after it passes*I'm stumped as to how to solve this. Please help me out. Also, please **DO NOT ROUND numbers until the very end**. If it's a number with multiple decimals, please round to the 7th or more place.

An airplane traveling with half the speed of

sound (i.e., 172 m/s) emits a sound of frequency 4.66 kHz.

1) At what frequency does a stationary listener hear the sound as the plane approaches?

Answer in units of kHz.

2) At what frequency does a stationary listener hear the sound after it passes?

Answer in units of kHz.

### 4 Answers | Add Yours

Nevermind, I got it. :)

The speed at which the sound travels remains same irrespective of the speed of an object, like an aeroplane, making the sound. Therefore, because of a phenomenon called Doppler effect the frequency of sound made by an object approaching a listener increases, while that be a receding object decreases.

The relation between the observed frequency of sound (f) and the emitted frequency (fo) is given by:

f = [v/(v + vs)]*fo

Where v is velocity of sound and vs is the velocity of source respective to the medium.

Also value of fo is given as 4.66 kHz.

It is given that speed of plane is half that of speed.

Therefore when the plane approaches the observer, it is like the intervening medium between the observer and the plane moving towards the observer making vs= -v/2

Therefore frequency of sound when the plane approaching the observer is given by:

f = [v/(v - v/2)]*fo = 2*fo = 2*4.66 = 9.32 kHz.

Similarly, after the plane passes the observer, it is like the intervening medium between the observer and the plane moving towards the observer making vs= v/2

Therefore frequency of sound after the plane passes the observer is given by:

f = [v/(v + v/2)]*fo = 2/3*fo = 2/3*4.66 = 3.10666667 kHz.

Let v be the speed of the sound and v' be the speed of the airplane. Let L be the distance of the airplane at ant time and the stationary observer. The first crest of sound wave of the airplane reaches the observer at L/v time . By this time of T0 seconds the airplanes comes towards the observer by T0*v'. So the next crest reaches observer at after an interval of [L-T0*v']/v. Or The successive time interval T of crests as herd by the observer is:

T = To+{L-To*v'}/v - L/v. Or

T = To{1-v'/v}. Therefore

1/T=f' = To*[1-v'/v]^(-1). Or

f' = f(1-v'/v)^(-1) = 4*66(1-1/2)^(-1) = (4.66)*2 = 9.32k Hrz.= 9320 hrz

When the plane goes away, the relation becomes f' =f(1+v'/v)^(-1) = f(1+1/2)^(-1) = 4.66KHz*(3/2)^(-1) = 4660*2/3 Herz= 9320/3 Herz = (3106+2/3)Herz.

a. the doppler equation for this is f' = f(1/(1-(Vs/V))) where f' = the apparent frequency, f = emitted frequency, Vs is the speed of the source, V is the speed of sound. This increases pitch as the wavelengths get shorter. The frequency goes up.

b. When the plan is moving away, the formula is f'=f(1/(1+(Vs/V))) this lengthens the wavelengths and decreses pitch. The frenquency decreases.