How Does Observer Velocity Affect Sound Wavefront Frequency?

  • Thread starter Thread starter Tanero
  • Start date Start date
  • Tags Tags
    Observer Sound
AI Thread Summary
The discussion focuses on how an observer's velocity affects the frequency of sound waves they perceive. When the observer moves away from the sound source at 40 m/s, the observed frequency decreases according to the Doppler effect, calculated using f' = (u/(u + vobs)) * fsource. The number of wavefronts arriving per unit time, or dN/dt, corresponds to the observed frequency. There is some confusion about the relevance of the distance from the source and whether additional information about the observer's initial state is needed. Overall, the key takeaway is that the frequency observed changes due to the observer's motion, impacting the rate of wavefronts detected.
Tanero
Messages
21
Reaction score
0

Homework Statement


A source of sound emits waves at a frequency f 450 Hz. An observer is located at a distance d 150 m from the source. If the observer is moving away from the source at a velocity vobs 40 m/s, how does the number of wavefronts change with time? dN/dt ? (in Hz)


Homework Equations


λ=u/f
N=d/λ
u - speed of sound


The Attempt at a Solution


with moving observer my frequency changes. f'=(u/u+vobs)*fsource
λ'=u/(u/u+vobs)*fsource)=(u+vobs)/f

in time Δt, ΔN=(d+vobsΔt)/λ' - d/λ'

then I find a limit for Δt>0 lim ΔN/Δt= dN/dt

Am I on the right way?
 
Physics news on Phys.org
1 wavefront arrives per cycle.
The number of wavefronts that arrive per unit time is the number of cycles per second - what quantity is measured in cycles per second?
 
  • Like
Likes 1 person
Simon Bridge said:
1 wavefront arrives per cycle.
The number of wavefronts that arrive per unit time is the number of cycles per second - what quantity is measured in cycles per second?

Just frequency? No need of all that ΔN/Δt calcs?

Then, if observer is receding f' = u/[(u+vobs)*fsource]
Is it? dN/dt = f' . Is that what you mean?
 
The number of wavefronts observed per second is the frequency of the sound observed :)

It's an odd way of asking for it though - and what does the distance to the source have to do with it?
It sounds to me like there may be some information missing from the problem statement.
Like if the observer was originally stationary and at t=0 suddenly starts moving ... then the frequency would change with time.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top