missed one lecture in physics class and never understood how to determin if plus or minus...you where the only channel i found that made it so clear and taught it just like my physicas teach did. Thank you
You know the ± on the top, is it the same sort of logic as with the symbol on the bottom? So, if the object in question is moving towards the observer, one would expect a higher frequency. And to get a higher frequency, you'd want a higher numerator. In other words, would it just be the opposite of the sign on the bottom?
If someone on a stage is addressing a crowd with a megaphone, based on the Earth's rotation if the crowd is moving into the sound waves at up to 460m per second or away from them, should there be a tonal difference in the sound wave the crowd would hear?
Michel van Biezen Allow me to give you 2 examples where this may not be the case. #1**- As Earth orbits the Sun at 66,000 km/h we are moving towards stars and away from these same stars given the season. Proof of Earth's orbit would be in seasonal doppler shifts due to our orbit. Do we detect seasonal shifts in starlight due to Earth's orbit? #2- As someone speaks do the sound-waves fire out like a projectile where if the waves hit a listener moving with or against the rotation, the listener is travelling faster or slower through the waves? Would this create a tonal difference in the waves?
Michel van Biezen Perfect and thank you. For #1, where is the data to show we are looking at doppler starlight effects from our yearly orbit cycle around and the Sun and not the daily perceived motion of stars which in a Geocentric model we would have a daily cycle of doppler starlight shifts as well? This would end the debate if you have actual data and proof of a yearly cycle of uniform doppler starlight shifts in the direction of our orbit and going away from stars. Does this data exist?
Sir I have a doubt: assuming that the speed of sound waves in air is 340m/s; if the wave is produced by a car going 20 m/s, will the wave travel at 360 m/s?
There is another easy way of doing this instead of the formula at 3:53. find the differential λ using the formula f = v/λ. So 10 m/s divided by 500 = 0.02 then 0.68m - 0.02m = 0.66m Now f = 340 m/s divided by 0.66 = 515.15Hz
Sir i have question for u 1)a source of sound travels towards a stationary object. The frequency of sound heard by the observer is 25% more the actual frequency. If speed of sound is v, that of source. Is what?
The basic wave equation is: velocity = frequency * wavelength or frequency = velocity / wavelength If you check the units you will get (1/sec) = Hz The interpretation is waves per second, or cycles per second, or oscillations per second.
Thank You. Love your videos. Now I know where to learn more about physics, after I drop from advanced physics. Is it better to solve a physics problem using units as your guide or should stick to the formulas?
WWadventure There are occasions when you will use one method and times when you'll use the other method. There are no absolutes to the problem solving technique
This guy deserves teacher of the year, Every year. Better than my lectures!! I understand this fully now. Explained it nice and simple. Thank you!!!!
the best physic teacher i know so far!!!! your the best sir thx alottttttt
missed one lecture in physics class and never understood how to determin if plus or minus...you where the only channel i found that made it so clear and taught it just like my physicas teach did. Thank you
I needed this, thanks for this excellent explanation. This channel is very reliable.
You are welcome.
Hahaha, brilliant sound demonstration on 3:05-3:10.
mmMMMMMMmmmmm doppler
You know the ± on the top, is it the same sort of logic as with the symbol on the bottom? So, if the object in question is moving towards the observer, one would expect a higher frequency. And to get a higher frequency, you'd want a higher numerator. In other words, would it just be the opposite of the sign on the bottom?
yes
thank you so much for clarifying this equation! I was so confused until I saw your video!
Excellent demonstration, thanks.
If someone on a stage is addressing a crowd with a megaphone, based on the Earth's rotation if the crowd is moving into the sound waves at up to 460m per second or away from them, should there be a tonal difference in the sound wave the crowd would hear?
No because everything moves with the rotating Earth.
Michel van Biezen
Allow me to give you 2 examples where this may not be the case.
#1**- As Earth orbits the Sun at 66,000 km/h we are moving towards stars and away from these same stars given the season. Proof of Earth's orbit would be in seasonal doppler shifts due to our orbit. Do we detect seasonal shifts in starlight due to Earth's orbit?
#2- As someone speaks do the sound-waves fire out like a projectile where if the waves hit a listener moving with or against the rotation, the listener is travelling faster or slower through the waves? Would this create a tonal difference in the waves?
#1 Yes
#2 No (They are very different situation)
Michel van Biezen
Perfect and thank you.
For #1, where is the data to show we are looking at doppler starlight effects from our yearly orbit cycle around and the Sun and not the daily perceived motion of stars which in a Geocentric model we would have a daily cycle of doppler starlight shifts as well?
This would end the debate if you have actual data and proof of a yearly cycle of uniform doppler starlight shifts in the direction of our orbit and going away from stars.
Does this data exist?
Sir I have a doubt: assuming that the speed of sound waves in air is 340m/s; if the wave is produced by a car going 20 m/s, will the wave travel at 360 m/s?
No, the wave will still travel at 340 m/sec, but the frequency and wavelength of the waves will change.
great lecture as always and also nice sound effects 3:03 lol
Glad you like them! 😂
does this equation keep sense if source is going at same velocity or faster than speed of sound ?
For that you'll have to look at the videos regarding the sonic boom in the same playlist.
Thanks great video
There is another easy way of doing this instead of the formula at 3:53.
find the differential λ using the formula f = v/λ. So 10 m/s divided by 500 = 0.02
then 0.68m - 0.02m = 0.66m
Now f = 340 m/s divided by 0.66 = 515.15Hz
great vid. now i understand the equation
Sir i have question for u
1)a source of sound travels towards a stationary object. The frequency of sound heard by the observer is 25% more the actual frequency. If speed of sound is v, that of source. Is what?
v(source) = (1/5) v f(obs) = f(source) x ((v)/(v - v(source))
What is frequency regarding waves? The definition of frequency is events/time. How does that fit in with waves?
The basic wave equation is:
velocity = frequency * wavelength
or frequency = velocity / wavelength
If you check the units you will get (1/sec) = Hz
The interpretation is waves per second, or cycles per second, or oscillations per second.
Thank You. Love your videos. Now I know where to learn more about physics, after I drop from advanced physics. Is it better to solve a physics problem using units as your guide or should stick to the formulas?
WWadventure
There are occasions when you will use one method and times when you'll use the other method. There are no absolutes to the problem solving technique
Thanks, much appreciated in 2018
we use the same model of calculator which you used in this video in India, sir. Casio is very famous
Love you man thanks alot!!!
thank you
You're welcome
Thanks
can u pls prove the equation
I'll have to add that to the list of videos to do. Thanks for the suggestion.
awesome!
im gonna write MMMMmmmmmm in my notes cause it works XD
"mmmmhhhhhhmmmmmmmm''
NNNNNNNNNNNNNNNNNNNNnnnnnnnnnnnnnnnnnnnnnnnnn
we use the same model of calculator which you used in this video in India, sir. Casio is very famous