# Doppler effect – problems and solutions

Doppler effect – problems and solutions

1.

(1) an observer moving toward the stationery source

(2) source moving toward the stationary observer

(3) observer and source approach each other

(4) observer and source are moving at the same speed

If the pitch heard is higher than that of the emitted source frequency, then which statement above are correct :

A. (1), (2) and (3)

B. (1), (2), (3) and (4)

C. (1) and (3)

D. (1) and (4)

E. (2) and (4)

Solution

The equation of the Doppler effect :

Sign rule :

The sound speed (v) always positive

The observer speed (vobs) is positive if observer moving toward the source of the sound

The observer speed (vobs) is negative if the observer moving away from the source of the sound

The source speed (vsource) is positive if the source of the sound moving away from the observer

The source speed (vsource) is negative if the source of the sound moving toward the observer

The observer speed (vobs) = 0 if an observer at rest

The source speed (vsource) = 0 if source at rest

For example :

The observer speed (vobs) = 60 m/s, if observer at rest then vobs = 0

The source speed (vsource) = 40 m/s, if the source of the sound at rest then vsource = 0

The sound speed (v) = 340 m/s

The frequency of sound (f) = 1000 Hertz

An observer moving toward the stationery source

The observer speed (vobs) is positive if observer moving toward the source of the sound

The source of the sound at rest so vsource = 0

Source moving toward the stationary observer

The source speed (vsource) is negative if the source of the sound moving toward an observer

Observer at rest so (vobs) = 0

Observer and source approach each other

The observer speed (vobs) is positive if observer moving toward the source of the sound

The source speed (vsource) is negative if the source of the sound moving toward the observer

Observer and source are moving at the same speed

If the source of the sound and observer moves at the same speed then no Doppler effect occurs.

2. An observer at rest near the source of the sound of frequency 684 Hz. Another the source of the sound of 676 Hz moving toward the observer at 2 n/s. If the speed of the sound waves in air is 340 m/s, then what is the beat frequency heard by the observer.

Known :

The frequency of the source of the sound 1 (f1) = 684 Hz (rest)

The frequency of the source of the sound 2 (f2) = 676 Hz (move)

The speed of the source of the sound 2 (v2) = 2 m/s (moving toward the observer)

The speed of the source of the sound waves in air (v) = 340 m/s

Wanted: The beat frequency heard by the observer

Solution :

The equation of the Doppler effect :

Sign rule :

The sound speed (v) always positive

The observer speed (vobs) is positive if observer moving toward the source of the sound

The observer speed (vobs) is negative if observer moving away from the source of the sound

The source speed (vsource) is positive if the source of the sound moving away from the observer

The source speed (vsource) is negative if the source of the sound moving toward the observer

The observer speed (vobs) = 0 if an observer at rest

The source speed (vsource) = 0 if source at rest

The beat frequency heard by the observer = 684 Hz – 680 Hz = 4 Hz.

3. A source of sound moving toward the stationary observer at 20 m/s. The frequency of the source of the sound = 380 Hz. The speed of the sound waves in air = 400 m s-1. What is the frequency of the sound waves heard by the observer?

Known :

The speed of the source of the sound (vsource) = 20 m/s

The speed of observer (vp) = 0

The frequency of the source of the sound (f) = 380 Hz

The speed of the source of the sound waves (v) = 400 m s-1

Wanted: The frequency of the sound waves heard by the observer

4. Car A moves at 72 km/h and car B moves at 90 km/h, approach each other. Car A honked with a frequency of 650 Hz. If the speed of the sound waves in air is 350 m/s, then what is the frequency of sound heard by the driver of car B from car A.

Known :

The speed of car A (vA) = 72 km/h = 20 m/s, approach car B

The speed of car B (vB) = 90 km/h = 25 m/s, approach car A

The frequency of the sound of car A (fA) = 650 Hz

The speed of the sound waves in air (v) = 350 m/s

Wanted: Frequency of sound heard by the driver of the car B from car A

Solution :

The sound speed (v) always positive

The observer speed (vobs) is positive if observer moving toward the source of the sound

The observer speed (vobs) is negative if the observer moving away from the source of the sound

The source speed (vsource) is positive if the source of the sound moving away from the observer

The source speed (vsource) is negative if the source of the sound moving toward the observer

The observer speed (vobs) = 0 if an observer at rest

The source speed (vsource) = 0 if source at rest

5. A source of sound moves at 10 m/s approach a stationary observer. The frequency of the source of the sound is 380 Hz and the speed of the sound waves in air is 400 m/s What is the frequency of the sound waves heard by the observer.

Known :

The speed of the source of the sound (vs) = 20 m/s

The speed of observer (vp) = 0

The frequency of the source of the sound (f) = 380 Hz

The speed of the sound waves in air (v) = 400 m/s

Wanted: The frequency of the sound waves heard by the observer

Solution :

The sound speed (v) always positive

The observer speed (vobs) is positive if observer moving toward the source of the sound

The observer speed (vobs) is negative if the observer moving away from the source of the sound

The source speed (vsource) is positive if the source of the sound moving away from observer

The source speed (vsource) is negative if the source of the sound moving toward the observer

The observer speed (vobs) = 0 if an observer at rest

The source speed (vsource) = 0 if source at rest

6. A car moving toward a stationary observer that emits 490 Hz sound wave. The beat frequency heard is 10 Hz. If the speed of the sound waves in air is 340 m/s, what is the speed of the car?

Known :

The frequency of sound (f) = 490 Hertz

The speed of the sound waves in air (v) = 340 m/s

The observer approaches the sound source so that the frequency of sound heard is greater than the frequency of the sound source. The frequency of sound = 490 Hertz and the beat frequency = 10 Hertz so that the frequency of the sound heard by an observer (f’) = 500 Hertz.

Wanted: the speed of the car

Solution :

The equation of the Doppler effect :

Sign rule :

The sound speed (v) always positive

The observer speed (vobs) is positive if observer moving toward the source of the sound

The observer speed (vobs) is negative if the observer moving away from the source of the sound

The source speed (vsource) is positive if the source of the sound moving away from the observer

The source speed (vsource) is negative if the source of the sound moving toward the observer

The observer speed (vobs) = 0 if an observer at rest

The source speed (vsource) = 0 if source at rest

The speed of the car is 6.9 m/s.

7. The police car that was ringing a 930 Hz siren chased after someone who ran away on a motorcycle with a speed of 72 km.jam-1. The speed of police cars reaches 108 km.hour-1. If the speed of sound in the air is 340 m.s-1, then the frequency of siren sounds heard by motorcyclists is …

Solution :

Sign rule :

The sound speed (v) always positive

The observer speed (vobs) is positive if observer moving toward the source of the sound

The observer speed (vobs) is negative if the observer moving away from the source of the sound

The source speed (vsource) is positive if the source of the sound moving away from the observer

The source speed (vsource) is negative if the source of the sound moving toward the observer

The observer speed (vobs) = 0 if an observer at rest

The source speed (vsource) = 0 if source at rest

Known :

The frequency of the source of sound (f) = 930 Hz

The speed of observer (vp) = 72 km.hour-1 = 72 (1000 meters) / 3600 (seconds) = 72,000/3600 meters/second = 20 m/s = -20 m.s-1

The speed of the source of sound (vsource) = 108 km.hour-1 = 108 (1000 meters) / (3600 seconds) = 108,000 / 3600 meters/second = 30 m/s = -30 m.s-1

The speed of the source of the sound waves (v) = 340 m.s-1

Wanted : The frequency of the sound waves heard by an observer (f’)

Solution :

The equation of the Doppler effect :

8. An ambulance moves at 72 km.hour-1 while sounding a siren with a frequency of 1500 Hz. Motorcyclists move at speeds of 20 m.s-1 in opposite directions with ambulances. If the speed of sound in air is 340 m.s-1, then the ratio of frequencies heard by motorcyclists when approaching and away from the ambulance is …

Known :

The frequency of the source of sound (f) = 1500 Hz

The speed of observer (vp) = 20 m/s

The speed of the source of sound (vsource) = 72 km/hour = 72 (1000 meters) / 3600 seconds = 72,000/3600 meters/seconds = 20 m/s

The speed of the source of the sound waves (v) = 340 m/s

Wanted: The ratio of frequencies heard by motorcyclists when approaching and away from the ambulance

Solution :

The equation of the Doppler effect :

Frequencies heard by motorcyclists approaching ambulances

Both are in opposite directions so that when the motorcycle approaches the ambulance car, the two approach each other. vp is positive if the listener approaches the sound source and vs is negative if the sound source approaches the listener.

Frequencies are heard by motorcyclists as they move away from the ambulance

Both are opposite direction so that when the motorcycle away from the ambulance, both of them away from each other. vp is negative if the listener is away from the sound source and vs is positive if the sound source is away from the listener.

The ratio of frequencies heard by motorcyclists when approaching and away from the ambulance

1. What is the Doppler effect?
• Answer: The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source.
2. How does the Doppler effect manifest in sound waves?
• Answer: When a sound source approaches an observer, the observer perceives a higher frequency (or pitch) than when the source is stationary. Conversely, as the sound source moves away from the observer, the perceived frequency is lower.
3. What is the difference between redshift and blueshift in terms of the Doppler effect?
• Answer: In terms of electromagnetic waves (like light), redshift refers to the shift of light towards longer wavelengths (or lower frequencies) when a source is moving away from an observer. Blueshift, on the other hand, refers to the shift of light towards shorter wavelengths (or higher frequencies) when the source is moving towards the observer.
4. How does the Doppler effect apply to galaxies and their movement in the universe?
• Answer: Many galaxies exhibit redshift, indicating they are moving away from us. This observation forms a key piece of evidence for the expansion of the universe.
5. What role does the Doppler effect play in weather radar technology?
• Answer: Doppler radar measures the change in frequency of the reflected radar signal caused by moving objects, like raindrops or hailstones. This allows meteorologists to detect the motion of precipitation and gauge wind speed and direction.
6. Is the Doppler effect observed only in sound and light waves?
• Answer: No, the Doppler effect can be observed in any type of wave, be it sound, light, or other electromagnetic radiation. It simply requires relative motion between the source and the observer.
7. How does the relative speed between the source and the observer affect the magnitude of the Doppler effect?
• Answer: The greater the relative speed between the source and the observer, the more pronounced the Doppler shift. For instance, a fast-moving ambulance will produce a more noticeable change in pitch as it passes by compared to a slowly moving one.
8. Why do we not notice a Doppler shift in the light of a car’s headlights as it approaches us at night?
• Answer: While the Doppler effect does occur for the car’s headlights, the speed of the car is much too small compared to the speed of light. Hence, the Doppler shift in frequency/wavelength of the light is minuscule and not detectable by the human eye.
9. How do animals like bats utilize the Doppler effect?
• Answer: Bats use echolocation, emitting sound waves and listening to the echoes to locate and catch prey. The Doppler effect plays a role when the bat or its prey is moving, causing a shift in the frequency of the echoed sound waves, which the bat can detect and use to judge motion and distance.
10. How does the Doppler effect relate to the siren of an emergency vehicle?
• Answer: As an emergency vehicle with a siren approaches, the sound waves get compressed, leading to a higher frequency or pitch. As the vehicle passes and moves away, the sound waves get stretched, resulting in a lower frequency or pitch. This is the Doppler effect in action, and it’s why the siren sounds different as the vehicle approaches and then recedes.