If the wind were blowing the other direction, would you subtract from top and bottom? All in all does adding or subtracting wind speed depend on the wind relative to the observer?
I don´t quite get the intuition on why the wind makes the frequency lower, in this case. It makes sense looking at the math, but leaving it aside I would think that the frequency was higher, because the wind is only letting the waves travel faster and reach the observer faster (unless we assume that the wind is also slowing down the observer. If that´s the case, it makes sense) Excellent video as always
+ProjectD100 remember [ velocity = wavelength*frequency] here source frequency is constant 500Hz ,so when velocity increases due to wind speed.The effective or observed wavelength must increase so the observed frequency is low as its inversely proportional to wavelength.
So wouldn’t that mean wind only has an effect on objects in motion? Based on this equation, if you have two stationary objects, the numerator and denominator would always equal 1; even if the wind was blowing 1000 mph. 500 x 1 = no change in frequency. What am I missing?
Excelent Videos. I follow a little bit different approach; the velocity of observer and the velocity of the source in the equation are relative to the air. So, the velocity of the source relative to the air is 20 - 10 = 10 m/s getting close to the observer (so use a negative in the denominator); the velocity of the observer relative to the air is 5 + 10 = 15 m/s getting close to the source (so use a positive in the numerator)...then fob = (500)(340+15) / (340 - 10) = 537.88 Hz...same result, different approach.
This is wrong, the wavelength of the wave as well as the speed of the wave changes, however the frequency observed remains constant (for the wind effect).
Wrong :) - Doppler effect if the source and the observer are not moving relative to each other the frequency is same if they move together or if the medium move same again for them
If the source and observer move relative to each other then the doppler shift term cancels to 1. If the medium moves and they do not move relative to each other, the multiplicative term simplifies to 1. Hence, you are incorrect.
You are the greatest Physics lecturer I have ever seen!
Way to answer an idle question within seconds. Thank you.
Thank you.
Thanks a lot sir... u really helped me to breeze through this concept
does the wind not appear on the bottom as negative since it's making the move towards the observer and increasing frequency?
If the wind were blowing the other direction, would you subtract from top and bottom? All in all does adding or subtracting wind speed depend on the wind relative to the observer?
I don´t quite get the intuition on why the wind makes the frequency lower, in this case. It makes sense looking at the math, but leaving it aside I would think that the frequency was higher, because the wind is only letting the waves travel faster and reach the observer faster (unless we assume that the wind is also slowing down the observer. If that´s the case, it makes sense)
Excellent video as always
+ProjectD100 remember [ velocity = wavelength*frequency] here source frequency is constant 500Hz ,so when velocity increases due to wind speed.The effective or observed wavelength must increase so the observed frequency is low as its inversely proportional to wavelength.
So wouldn’t that mean wind only has an effect on objects in motion? Based on this equation, if you have two stationary objects, the numerator and denominator would always equal 1; even if the wind was blowing 1000 mph. 500 x 1 = no change in frequency. What am I missing?
Indeed, the wind does not affect the doppler shift. Only the motion of the source and the observer.
completely satisfied thanks a lot cleared all my doubts
Excelent Videos.
I follow a little bit different approach; the velocity of observer and the velocity of the source in the equation are relative to the air. So, the velocity of the source relative to the air is 20 - 10 = 10 m/s getting close to the observer (so use a negative in the denominator); the velocity of the observer relative to the air is 5 + 10 = 15 m/s getting close to the source (so use a positive in the numerator)...then fob = (500)(340+15) / (340 - 10) = 537.88 Hz...same result, different approach.
+Aníbal C. Ripoll R.
Good input.
why not negative in denominator with wind velocity?
The velocity of the wind only has the effect of the apparent increase of the speed of the sound in air
Thank you Sir!
i think the wind speed is negative at the bottom
The equation in the video is correct. Thank you for checking.
This is wrong, the wavelength of the wave as well as the speed of the wave changes, however the frequency observed remains constant (for the wind effect).
Actually I retract my comment. If the source is moving the wind does have an effect indeed. Hats off to you sir..
Nice
Glad you liked it. 🙂
Wrong :) - Doppler effect if the source and the observer are not moving relative to each other the frequency is same if they move together or if the medium move same again for them
If the source and observer move relative to each other then the doppler shift term cancels to 1. If the medium moves and they do not move relative to each other, the multiplicative term simplifies to 1. Hence, you are incorrect.
nice vedios... but why wont you just remove the flawed vedios altogether
+Anjum Nazir Some people like to go find the error / flaw. It can be a learning experience.