How to derive the doppler effect formula for sound

The Doppler shift formula allows you to determine the perceived frequency of a sound source, but how is it determined and derived. This video goes through the derivation on how it is worked out.
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[@tastypie2276]

Thank you so much, Sir! It really helped to understand where the formula came from.

[@robertsimpson4412]

Nice video but I think you need to add some more explanation about why you keep the wavelength constant in the 2nd case. It's quite common to see students mix up the two cases.

[@jakeaustria5445]

Thank you so much for this. I need it for a contest.

[@felistus3439]

you the man I'm ready for my test tomorrow morning❤

[@alfinbarouq55]

Thank you sir, it really helps me out

[@sebcodestheweb9385]

Now this is a teacher I follow

[@joshuachama8]

This was helpful. Thank you.

[@PhysicsHigh]

Glad it was helpful!

[@diatribeeverything]

Great video. You only lost me at one point. Can you please explain how you combined your two equations, in white, because it looks like you didn't set them equal to each other... as that would cancel frequency. It's like you only multiplied a part of them. Can you give me an example of this move that is simpler? Thanks!
Awesome shirt btw!

[@PhysicsHigh]

Thank for the feedback.
In answer to your question...It's not so much that they add numerically. Instead the final formula can show both situations at the same time. In fact, the two are multiplied, as you suggest
Think of this way....
Let's make EQo is the equation where observer is moving and EQs is the equation the source is moving.
f' = f(EQo) but the source is moving too f'' = f'(EQs)
So you have f'' = f (EQo)(EQs)
Hope that makes sense

[@andrewc3937]

@@PhysicsHigh This will make sense if you use f'' instead of f' in your second equation.

[@user-jh3kz7dp2z]

@@andrewc3937 well that's kinda not how it works because both formulas are applied simultaneously

[@bernardolinares4162]

@@PhysicsHigh I understand how the result is achieved mathematically, but I don't see the motivation behind combining the equations. I'm teaching this in high school, and I have a physics degree, but every once in a while I need a refresher. From my understanding, the observed frequency with respect to the observer (f') ends up being the observed frequency with respect to the source (f''). From there the solution is obvious, algebraically. However, I want to come up with a good and simple explanation as to why f'' is the one that is important. This is clearly a significant result because it can be applied in all situations involving the Doppler effect.

[@MrYahya0101]

Combining them in this way: f' = v/wavelength' = v/(v*T - vs*Ts - v'*T') gives => f' = fs*(v+v')/(v-vs). In other words, you have to account for the observer moving and the source moving using f' = v/wavelength'. As you said, multiplying them doesn't give the final equation.

[@georgekordas6184]

Make the same analysis for both. Not beginning from for λ' in the first case and from the T' in the second

[@mirtacota]

Thank u, excellent explanation!

[@G9Gupta]

One question: @7:33
Can we just combine the formulas like this?
I think we played it save as results matches but ideally it should be solved taking mothion of both observer and source
But again this will ha0pen if superposition principle can be applied here because of linear nature of equations
But ya thats a good way to solve equations if you good intuition thanks 😅

[@robertobaldizon6010]

Thank you very much, Sir. Very clear.

[@MrYahya0101]

I don't see why in one part it's f' = v/wavelength' and in the second it's f' = (v + v')/wavelength (I'll use v to denote velocity of the wave). Basically. the second equation indicates there will be no change in the perceived wavelength, just a change in the velocity. And I'm not sure if that's accurate. If you use f' = v/wavelength' for both, you still get the same equations. For the first one f' = v/ (v*T - vs*T) (which is what you have) and for the second one f' = v/(v*T - v'*T') => f' = fs*(1+v'/v).

[@ayasaki.pb_787]

wavelength was constant in second case because the source stationary. The moving observer preceive faster wave speed over the same distance.

[@sharanbalajimanickamoorthy3639]

Amazing

[@preparingforundergraduatep4256]

why process frequency is = velocity of wave / process lambda (why velocity of wave?)

[@anandapatmanabhansu]

Thankyou sir

[@lvenE-kx4dq]

I have a question about converting Vsource *time /λ to Vsource / Vwave, as the time might not equal the time taken for the wave to move for 1 period, so can it be converted?

[@verysimplephysicsinthamil1436]

Sir,your class is fantastic..sir I am a taecher of physics teaching class 11 and 12 ,I am also using digital pentablet for recording videos for presentation,could you tell me what are all the hardwares and software you use for recording...

[@stevecorlew7187]

Would be interested in what app you use. Would like to hear the doppler effect on geometric shape, speculation an elipse. Great video 👍🏽👍🏽

[@truth30]

why we say t =T in d=vst because it is time taken for source to cover some d,not for wave to cover lamda

[@kashifuddinahmed8383]

I used to love doing physics in high school but it's been too long now don't remember much heck even my basic math is messed up.

[@bernardwhipps7558]

I should have paid more attention in high school

[@TheFixer693]

I do not understand why you divided for the final stage, the coefficients of the frequency of the wave.

[@mitchlg2005]

First