The doppler effect
The Doppler Effect is what happens when an object producing sound moves closer to, or away from, another object. The object observing the sound will receive a higher frequency if the object producing the sound is moving closer, and a lower frequency if it's moving away. This is called the perceived frequency, and is produced by the sound waves getting pushed together (higher perceived frequency) and the waves becoming more spaced out (lower perceived frequency). The Doppler Effect can also occur when the observer is traveling as well. Here is the equation:
For objects moving closer to each other
f'= (V+Vo) (f)
(V-Vs)
For objects moving farther away from each other
f'= (V-Vo) (f)
(V+Vs)
f'= perceived frequency
V= velocity of sound
Vo= velocity of the observer
Vs= velocity of the source of sound
Example (objects moving closer together)
f'= ?
f= 680Hz
V= 343m/s
Vo= 10m/s
Vs= 30m/s
f'= (343m/s+10m/s) (680Hz) = f'= (353m/s) (680Hz) = f'= (1.128)(680Hz) = f'= 767Hz
(343m/s-30m/s) (313m/s)
The perceived frequency for this problem should be higher than the original frequency because the objects are getting closer together, therefore pushing the sound waves together.
For objects moving closer to each other
f'= (V+Vo) (f)
(V-Vs)
For objects moving farther away from each other
f'= (V-Vo) (f)
(V+Vs)
f'= perceived frequency
V= velocity of sound
Vo= velocity of the observer
Vs= velocity of the source of sound
Example (objects moving closer together)
f'= ?
f= 680Hz
V= 343m/s
Vo= 10m/s
Vs= 30m/s
f'= (343m/s+10m/s) (680Hz) = f'= (353m/s) (680Hz) = f'= (1.128)(680Hz) = f'= 767Hz
(343m/s-30m/s) (313m/s)
The perceived frequency for this problem should be higher than the original frequency because the objects are getting closer together, therefore pushing the sound waves together.
Sound interference
Sound interference is what happens when two or more sound sources produce sound waves. There are two types: constructive and destructive. If the rarefaction and compression areas in sound waves line up, then the sound that is observed will be higher in amplitude (constructive). If the rarefactions and compressions of one sound wave oppose another, so that they look "opposite", then the resulting perceived sound will be lower in amplitude and can result in no perceived sound at all (deconstructive). The two sounds "cancel each other out". Constructive interference occurs every whole wavelength and destructive interference occurs every half wavelength. The equations to find sound interference are as follows:
n= d2-d1
wavelength
n= a unitless number. If the number is a whole number (3, 5, ect.), or close to one, then the interference is constructive. If the resulting number is a half number (1.5, 2.5, ect.), or close to one, then the interference is destructive).
d1= distance 1. This is the distance an object is away from one source of sound.
d2= distance 2. This is the distance an object is away from the other source of sound.
wavelength (lambda)= the measurement of the peak of one sound wave, to the peak of the adjacent sound wave.
The equation to find lambda is as follows:
lambda= V/f
Example
Constructive or destructive?
d1=10m
d2= 20m
V= 343m/s
f= 340Hz
lambda= 343m/s
340Hz
lambda= 1.01m
n= 20m-10m
1.01m
n= 9.9= constructive
n= d2-d1
wavelength
n= a unitless number. If the number is a whole number (3, 5, ect.), or close to one, then the interference is constructive. If the resulting number is a half number (1.5, 2.5, ect.), or close to one, then the interference is destructive).
d1= distance 1. This is the distance an object is away from one source of sound.
d2= distance 2. This is the distance an object is away from the other source of sound.
wavelength (lambda)= the measurement of the peak of one sound wave, to the peak of the adjacent sound wave.
The equation to find lambda is as follows:
lambda= V/f
Example
Constructive or destructive?
d1=10m
d2= 20m
V= 343m/s
f= 340Hz
lambda= 343m/s
340Hz
lambda= 1.01m
n= 20m-10m
1.01m
n= 9.9= constructive