Asked by juanpro
A source of sound emits waves at a frequency 650 Hz. An observer is located at a distance 190 m from the source. Use m/s for the speed of sound.
(a) Assume completely still air. How many wavefronts (full waves) are there between the source and the observer?
unanswered
(b) If the observer is moving away from the source at a (radial) velocity m/s, how does the number of wavefronts found in part (a) change with time? For the answer, give the rate of change of , namely (in Hz)
unanswered
(c) By comparing the difference of the rate of wavefronts leaving and wavefronts entering the region between source and observer, calculate the frequency observed by the moving observer. (in Hz)
hint: how does the difference relate to the rate of change of you calculated in (b)?
unanswered
(a) Assume completely still air. How many wavefronts (full waves) are there between the source and the observer?
unanswered
(b) If the observer is moving away from the source at a (radial) velocity m/s, how does the number of wavefronts found in part (a) change with time? For the answer, give the rate of change of , namely (in Hz)
unanswered
(c) By comparing the difference of the rate of wavefronts leaving and wavefronts entering the region between source and observer, calculate the frequency observed by the moving observer. (in Hz)
hint: how does the difference relate to the rate of change of you calculated in (b)?
unanswered
Answers
Answered by
Anonymous
don't cheat
Answered by
jennifer
a) t=d/v = 0.5588 seconds
N=f*t = 650*0.5588= 363.235
b) i don't know it
c) f'= f* (1-(speed observer/speed sound))
N=f*t = 650*0.5588= 363.235
b) i don't know it
c) f'= f* (1-(speed observer/speed sound))
Answered by
Daoine
jennifer for b I was thinking if we know N = ft and we derive this wouldn't it be the same f? I mean dN/dt = d(f*t)/dt that results in f. The other idea I had was calculating the wavelenght (lambda = v/f) using the sound velocity and the given f and then knowing lambda use it to find the new f that would result in the derivative given
Answered by
elli
from a we have:
N=f.d/Vs
for b:
N=f.(d+v.t)/Vs
dN/dt=f.v/Vs
N=f.d/Vs
for b:
N=f.(d+v.t)/Vs
dN/dt=f.v/Vs
Answered by
elli
but, how did you do with D)???
Answered by
elli
so I risked one chance and got:
a) N=f.d/Vs
b) dN/dt=f.Vobs/Vs
c) f'=f(1-Vobs/Vs)
and I thought for d:
f'=f.Vs/(Vs-Vwind)
but it's not right, I'll assume f' have to be bigger since, Vwind goes from source to observer.. but well, any suggestions?
a) N=f.d/Vs
b) dN/dt=f.Vobs/Vs
c) f'=f(1-Vobs/Vs)
and I thought for d:
f'=f.Vs/(Vs-Vwind)
but it's not right, I'll assume f' have to be bigger since, Vwind goes from source to observer.. but well, any suggestions?
Answered by
Daoine
I think that the fecuency remains the same in D. Because source and observer are at rest. By the same I mean that it could be 650 or the one you at c.
Answered by
Daoine
Look ellie you're right about b and c and d is 650 =)
Answered by
Anonymous
for D, there is no relative speed, so whatever is gained by the wind at source is lost at observer (or vice versa), so no change
Answered by
juanpro
Let us now assume that both source and observer are at rest, but wind blows at a constant speed v=20 m/s in the direction source towards observer. By comparing the difference of the rate of wavefronts
leaving and wavefronts entering the region between source and observer, calculate the observed frequency f
′? (in Hz)no answer iqual
leaving and wavefronts entering the region between source and observer, calculate the observed frequency f
′? (in Hz)no answer iqual
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