http://www.flickr.com/photos/54567186@N0…
Could you also mention what do these plus and minus signs mean and when to use which?
Thank you!
Could you also mention what do these plus and minus signs mean and when to use which?
Thank you!
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That's the equation for the Doppler effect. It describes the frequency the detector, ƒ', hears, according to the frequency of the source, ƒ, when the detector is moving at vd and when the source is moving at vs. Regular v is the speed of sound in air, roughly 343 m/s.
Basically, the signs are like that because when one is +, the other becomes −.
For example:
A source of 4 kHz sound waves travels at 38.1 m/s toward a detector that's moving 38.1 m/s toward the source. What is the frequency of the waves as they're received by the detector?
Let's say we have two objects, a source S and a detector D. We define ← as a negative direction, and → as a positive direction.
← →
D S
Because the detector moves toward the source (the → direction), we use + in the numerator, and since the source moves toward the detector (in the ← direction), we use the − in the denominator.
ƒ' = ƒ[(v + vd)/(v − vs)]
= (4000 Hz)[(343 m/s + 38.1 m/s)/(343 m/s − 38.1 m/s)]
= (4000 Hz)[(381.1 m/s)/(304.9 m/s)]
≈ 5000 Hz
= 5 kHz
Therefore, the detector hears 5 kHz due to the Doppler effect.
Basically, the signs are like that because when one is +, the other becomes −.
For example:
A source of 4 kHz sound waves travels at 38.1 m/s toward a detector that's moving 38.1 m/s toward the source. What is the frequency of the waves as they're received by the detector?
Let's say we have two objects, a source S and a detector D. We define ← as a negative direction, and → as a positive direction.
← →
D S
Because the detector moves toward the source (the → direction), we use + in the numerator, and since the source moves toward the detector (in the ← direction), we use the − in the denominator.
ƒ' = ƒ[(v + vd)/(v − vs)]
= (4000 Hz)[(343 m/s + 38.1 m/s)/(343 m/s − 38.1 m/s)]
= (4000 Hz)[(381.1 m/s)/(304.9 m/s)]
≈ 5000 Hz
= 5 kHz
Therefore, the detector hears 5 kHz due to the Doppler effect.