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Name: Neil Kyle O. Maniquis
Section: 10 - Copernicus
Date Submitted: November
18, 2021
WORKSHEET NO. 1
DOPPLER EFFECT
KEY IDEAS
Terms
DOPPLER EFFECT
Description
The apparent change in the frequency of a wave whenever there is
relative motion between a source of waves and an observer.
๏ƒ There is relative motion between two objects when the distance
between them is changing (either increasing or decreasing) in
time
๏ƒ The observed frequency will be higher than the actual frequency
when the distance between the source and observer is decreasing
(approaching)
๏ƒ The observed frequency will be lower than the actual frequency
when the distance between the source and the observer is
increasing in time (receding)
๏ƒ can be observed in all waves: mechanical or electromagnetic
Observed Frequency
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ค๐‘Ž๐‘ฃ๐‘’ ๐‘Ÿ๐‘’๐‘™๐‘Ž๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘ก๐‘œ ๐‘œ๐‘๐‘ ๐‘’๐‘Ÿ๐‘ฃ๐‘’๐‘Ÿ
๐‘“๐‘œ = ๐‘“๐‘  (
)
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ค๐‘Ž๐‘ฃ๐‘’ ๐‘Ÿ๐‘’๐‘™๐‘Ž๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘œ๐‘ข๐‘Ÿ๐‘๐‘’
Note on relative
velocity
Consider two objects, A and B, that are traveling in the same direction
along a straight path.
vB
vA
The velocity of car A relative to car B is defined as:
๐‘ฃ๐ด๐ต = ๐‘ฃ๐ด โˆ’ ๐‘ฃ๐ต
If car B is traveling in the opposite direction, then
๐‘ฃ๐ด๐ต = ๐‘ฃ๐ด โˆ’ (โˆ’๐‘ฃ๐ต ) = ๐‘ฃ๐ด + ๐‘ฃ๐ต
Bow wave
Occurs when the source of the wave is traveling at
a speed equal to the speed of the wave if
produces.
๏ƒ Also known as wave barrier
SHOCK WAVE
A shock wave is produced when an object travels at a speed greater
than the speed of the wave
๏ƒ the object need not be a source of
waves
๏ƒ it is formed by superposition of
wavefronts behind the object
๏ƒ for sound waves, the sound produced is
called the sonic boom
MACH NO.
The slope of the shock wave depends on the ratio of the speed of the
wave to the speed of the object.
sin ๐œƒ =
๐‘ฃ๐‘ค
1
=
๐‘ฃ๐‘œ
๐‘€
The Mach no. of a moving object is defined as the ratio of the speed of
the object to the speed of sound
๐‘€=
๐‘ฃ๐‘œ๐‘๐‘—๐‘’๐‘๐‘ก
๐‘ฃ๐‘ ๐‘œ๐‘ข๐‘›๐‘‘
๏ƒ Objects with M > 1 are traveling at supersonic speed. They
generate shock waves
๏ƒ Objects with M < 1 are traveling at subsonic speed. Doppler effect
can be observed in this case.
SAMPLE PROBLEMS
1. A railroad train is traveling at 25 m/s in still air. The frequency of the note emitted by the
locomotive whistle is 400 Hz. What is the frequency of the sound heard by a stationary listener
(a) in front of the locomotive? (b) behind the locomotive?
Solution: Draw a diagram showing the relative position of the source and the observer. Indicate the
velocities of the source, observer and the wave.
a. for the observer in front of the train (Observer A)
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ค๐‘Ž๐‘ฃ๐‘’ ๐‘Ÿ๐‘’๐‘™๐‘Ž๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘ก๐‘œ ๐‘œ๐‘๐‘ ๐‘’๐‘Ÿ๐‘ฃ๐‘’๐‘Ÿ
๐‘“๐ด = ๐‘“๐‘  (
)
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ค๐‘Ž๐‘ฃ๐‘’ ๐‘Ÿ๐‘’๐‘™๐‘Ž๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘œ๐‘ข๐‘Ÿ๐‘๐‘’
๐‘š
340 ๐‘  โˆ’ 0
๐‘ฃ๐‘ค โˆ’ ๐‘ฃ๐ด
๐‘“๐ด = ๐‘“๐‘  (
) = 400 ๐ป๐‘ง (
๐‘š
๐‘š) = 432 ๐ป๐‘ง
๐‘ฃ๐‘ค โˆ’ ๐‘ฃ๐‘ 
340 ๐‘  โˆ’ 25 ๐‘ 
Note: the wave and the source are traveling in the same direction.
b. for the observer behind the train (observer B)
๐‘š
340 โˆ’ 0
๐‘ฃ๐‘ค โˆ’ ๐‘ฃ๐ต
๐‘ 
๐‘“๐ต = ๐‘“๐‘  (
) = 400 ๐ป๐‘ง (
๐‘š
๐‘š) = 373 ๐ป๐‘ง
๐‘ฃ๐‘ค โˆ’ ๐‘ฃ๐‘ 
340 + 25
๐‘ 
๐‘ 
Note: in this case, the wave and the source are traveling in opposite directions.
2. A stationary motion detector sends sound waves of frequency 0.150 MHz toward a truck
approaching at a speed of 45 m/s. What is the frequency of the wave reflected back to the
detector?
Solution: To solve this problem, we break the situation into two parts:
First Part: the motion detector (source) sends waves toward the truck (observer).
The frequency of sound received by the truck is given by
๐‘ฃ๐‘ค + ๐‘ฃ๐‘‡
๐‘“๐‘œ = ๐‘“๐‘  (
)
๐‘ฃ๐‘ค
Second part: The truck (source) reflects the waves back to the motion detector.
The frequency reflected back to the detector is thus
๐‘ฃ๐‘ค
๐‘ฃ๐‘ค + ๐‘ฃ๐‘‡
๐‘ฃ๐‘ค
๐‘ฃ๐‘ค + ๐‘ฃ๐‘‡
340 + 45
๐‘“ โ€ฒ = ๐‘“๐‘œ (
)= (
)(
)= (
) = 0.150๐‘€๐ป๐‘ง (
) = 0.2 ๐‘€๐ป๐‘ง
๐‘ฃ๐‘ค โˆ’ ๐‘ฃ๐‘‡
๐‘ฃ๐‘ค
๐‘ฃ๐‘ค โˆ’ ๐‘ฃ๐‘‡
๐‘ฃ๐‘ค โˆ’ ๐‘ฃ๐‘‡
340 โˆ’ 45
EXERCISES
1. Five expanding wave fronts from a moving sound source are shown. The dots represent the
centers of the respective circular wave fronts, which is the location of the source when that wave
front was emitted. The frequency of the sound emitted by the source is constant.
a. Indicate on the figure the direction of motion of the source. Which sound wave front was
produced first (label it)? How do you know? Explain.
The labeled sound wave front is the first produced and this is known because it has the largest
diameter meaning it has been the first sound wave front produced. The explanation sounds
rather subjective but another reasoning for this is that, we know that given a point (source), if
an observer โ€˜s observance of the sounds is more frequent, then the sound wave is heading
towards the observer, which is observer B and since it is heading towards observer B, the point
nearest to observer B is the source of the newest circular wave front formed and the farthest
from it, is the source of the oldest circular wave.
b. Do the observers at locations A and B hear the same frequency of sound? If not, which one
hears a higher frequency? Explain.
No. Observer B hears a higher frequency because the speed of the wave is constant and it
can be seen that the wavefronts towards B are more compressed and since they move in the
same speed, more wavefronts will arrive at B before at A.
c. Assume that the sound wave you identified in part a as the first wavefront produced marks the
beginning of the sound. Do the observers at A and B first hear the sound at the same time? If
not, which one hears the sound first? Explain.
Based on the figure, the first point is equidistant to both observers A and B, and since it has
been stated that the objective frequency is the same, yes, they will hear the first sound at the
same time.
2. You are standing at x = 0 m, listening to a sound that is emitted at a frequency fo. At t = 0s, the
sound source is at x = 20 m and moving toward you at a STEADY 10 m/s. Draw a graph showing the
frequency you hear from t = 0s to t = 4s. ONLY the shape of the graph is important, not the numerical
value of f.
3. Two train whistles, A and B, each have a frequency of 392 Hz. A is stationary and B is moving
toward the right (away from A) at a speed of 35 m/s. A listener is between the two whistles and is
moving toward the right with a speed of 15 m/s.
a. What is the frequency from A as heard by the listener? //assuming still air
๐‘š
๐‘š
340 ๐‘  โˆ’ 15 ๐‘  )
392 ๐ป๐‘ง(
) = 374.71 ๐ป๐‘ง
๐‘š
340 ๐‘ 
b. What is the frequency from B as heard by the listener?
๐‘š
๐‘š
340 ๐‘  + 15 ๐‘ 
392 ๐ป๐‘ง(
๐‘š
๐‘š) = 371.09 ๐ป๐‘ง
340 ๐‘  + 35 ๐‘ 
4. A bat flies toward a wall at a speed of 10 m/s, emitting a steady sound of frequency 2000 Hz. At
what frequency does the bat hear the reflected sound?(Hint: Break this problem into two parts, first
with the bat as the source and the wall as the listener and then with the wall as the source and the
bat as the listener)
Let fโ€™ be the observed frequency of the sound and fโ€™โ€™ be the observed frequency of the reflected
sound.
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ค๐‘Ž๐‘ฃ๐‘’ ๐‘Ÿ๐‘’๐‘™๐‘Ž๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘ก๐‘œ ๐‘œ๐‘๐‘ ๐‘’๐‘Ÿ๐‘ฃ๐‘’๐‘Ÿ
) ; ๐‘œ๐‘๐‘ ๐‘’๐‘Ÿ๐‘ฃ๐‘’๐‘Ÿ = ๐‘ค๐‘Ž๐‘™๐‘™; ๐‘ ๐‘œ๐‘ข๐‘Ÿ๐‘๐‘’ = ๐‘๐‘Ž๐‘ก
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ค๐‘Ž๐‘ฃ๐‘’ ๐‘Ÿ๐‘’๐‘™๐‘Ž๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘œ๐‘ข๐‘Ÿ๐‘๐‘’
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ค๐‘Ž๐‘ฃ๐‘’ ๐‘Ÿ๐‘’๐‘™๐‘Ž๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘ก๐‘œ ๐‘œ๐‘๐‘ ๐‘’๐‘Ÿ๐‘ฃ๐‘’๐‘Ÿ
๐‘“ โ€ฒโ€ฒ = ๐‘“โ€ฒ (
) ; ๐‘œ๐‘๐‘ ๐‘’๐‘Ÿ๐‘ฃ๐‘’๐‘Ÿ = ๐‘๐‘Ž๐‘ก; ๐‘ ๐‘œ๐‘ข๐‘Ÿ๐‘๐‘’ = ๐‘ค๐‘Ž๐‘™๐‘™
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ค๐‘Ž๐‘ฃ๐‘’ ๐‘Ÿ๐‘’๐‘™๐‘Ž๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘œ๐‘ข๐‘Ÿ๐‘๐‘’
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’ + ๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘๐‘Ž๐‘ก
๐‘“ โ€ฒ = ๐‘“๐‘  (
) ; ๐‘“ โ€ฒโ€ฒ = ๐‘“โ€ฒ (
)
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’ โˆ’ ๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘๐‘Ž๐‘ก
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’ + ๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘๐‘Ž๐‘ก
๐‘“ โ€ฒโ€ฒ = ๐‘“๐‘  (
)(
)
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’ โˆ’ ๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘๐‘Ž๐‘ก
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’ + ๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘๐‘Ž๐‘ก
๐‘“ โ€ฒโ€ฒ = ๐‘“๐‘  (
)(
)
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’ โˆ’ ๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘๐‘Ž๐‘ก
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’ + ๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘๐‘Ž๐‘ก
๐‘“ โ€ฒโ€ฒ = ๐‘“๐‘  (
)
๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘ค๐‘Ž๐‘ฃ๐‘’ โˆ’ ๐‘ฃ๐‘’๐‘™๐‘œ๐‘๐‘–๐‘ก๐‘ฆ๐‘๐‘Ž๐‘ก
๐‘š
๐‘š
๐‘š
340 ๐‘  + 10 ๐‘ 
350 ๐‘ 
โ€ฒโ€ฒ
๐‘“ = (2000 ๐ป๐‘ง) (
๐‘š
๐‘š ) = 2000 ๐ป๐‘ง โˆ—
๐‘š = 2121.21 ๐ป๐‘ง
340 ๐‘  โˆ’ 10 ๐‘ 
330 ๐‘ 
๐‘“ โ€ฒ = ๐‘“๐‘  (