Question

1) A car is approaching a reflecting wall. A stationary observer behind the car hears a sound of frequency 790 Hz from the car horn and a sound of frequency 873 Hz from the wall.

(a) How fast is the car traveling?
.......... km/h

(b) What is the frequency of the car horn?
........... Hz

(c) What frequency does the car driver hear reflected from the wall?
............ Hz

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2) A bat flying toward an obstacle at 10 m/s emits brief, high-frequency sound pulses at a repetition frequency of 77 Hz. What is the time interval between the echo pulses heard by the bat?
.......... ms

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3) Using 340 m/s for the speed of sound in air, calculate the fundamental frequency for a 8.6 m organ pipe with the following characteristics.

(a) open at both ends
........... Hz

(b) closed at one end
........... Hz

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Answer 1

1. solution (similar problem)

A car is approaching a reflecting wall. A stationary observer behind the car hears a sound of frequency 745 Hz from the car horn and a sound of frequency 863 Hz from the wall.

1. How fast is the car traveling?

A. 25.2 m/s This is the answer. The first thing to do is answer question 2. Then determine what the difference in wavelength is and multiply that difference in wavelength by the original frequency.

2. The frequency of the car horn is in between the two frequencies heard. The lower frequency comes directly from the vehicle that is going away from the observer. The higher frequency comes from the wall which is a head of the vehicle. The frequency reaching the wall is the frequency leaving the wall. The observer and the wall have no relative motion so the frequency of the sound leaving the wall is the frequency received by the observer.

800 Hz This is the answer. The difference in the wave length between the original frequency and the wavelength of each of the observed frequencies is identical. The wavelength is calculated by dividing the the speed of sound by the frequency of the sound.


3. The frequency the car driver hears reflected from the wall is

931 Hz This is the answer. The sound leaving the wall has a frequency of 863 Hz which has a wave length of 0.397450753 meters and a cycle time of 0.001158749 seconds. The car travels 0.029166162 meters in that time. The wavelength is therefore shortened by that distance from the perspective of the driver of the car to 0.368284591 meters. dividing 343 by .368284591 gives 931 Hz.

2. solution

The velocity of sound in air is on the order of 340m/s.

For relatively slow sources, such as this, the Doppler shift is given by


f=f0*(1+(v source)/(v in medium))

f=77Hz*(1+10/340)

f=79026Hz

So the time interval is 1/f=0.012158seconds

If you're wondering why he hears himself at a higher frequency, it's because he's moving toward his pulses, so they're effectively scrunched together. To help imagine this, think of circles coming off of an object. Let those circles dissipate at a constant rate. Now if you move in one direction, let those pulses continue pulsing at the same rate. You'll see in your mind that on one side, the circles will be closer to each other, and on the other, they will be farther apart.

This is analogous to sound, in that when something is approaching you, it sounds higher, and if it's leaving you, it sounds deeper.

3.solution

For pipe closed at one end

f = v/4*l

therefore f = 340/4*8.6 = 9.88 m

For pipe open at both ends

f = v/2*l

therefore f = 340/2.8.6 =19.76