- #1
Charlie X
- 5
- 0
the formula of Doppler Effect
f = fs(v + vd)/(v - vs)
(v = speed of sound; d = detector; s = source)
Simply from this formula, it can be seen that vd has differenct effect on the frequency receiver from vs.
when vd or vs approaches the speed of sound, this difference is pronounced.
make v = 343 m/s, fs = 1 Hz, the the source and detector move towards each other
Case 1
vd = 340 m/s, vs = 0
f = 683/343 = 1.99 Hz
Case 2
vd = 0, vs = 340 m/s,
f = 343/3 = 114 Hz
why would there be a difference? Isn't the source and detector move relative to each other?
my teacher says it has sth to do with medium, but i still don't get it.
f = fs(v + vd)/(v - vs)
(v = speed of sound; d = detector; s = source)
Simply from this formula, it can be seen that vd has differenct effect on the frequency receiver from vs.
when vd or vs approaches the speed of sound, this difference is pronounced.
make v = 343 m/s, fs = 1 Hz, the the source and detector move towards each other
Case 1
vd = 340 m/s, vs = 0
f = 683/343 = 1.99 Hz
Case 2
vd = 0, vs = 340 m/s,
f = 343/3 = 114 Hz
why would there be a difference? Isn't the source and detector move relative to each other?
my teacher says it has sth to do with medium, but i still don't get it.