The Doppler Formula

Transcript

0:00Introduction

Hello everyone! Let's start with a puzzle. You stand on a railway platform, and a train rushes past, blowing its whistle. As it goes by, the pitch drops: high, then low. But the driver, sitting next to the whistle, hears one steady note. So who is right, you or the driver? Hold on to that question. By the end of this video, you will have the answer, and you will even find the speed of the train. This is the Doppler effect, and it comes up often in JEE, so let's begin.

0:32Moving source: the waves crowd ahead

So, look at the figure. The orange dot, capital S, is the source of sound, and the arrow shows it moving right with speed v sub s, the v with a small s below it. Each circle is a wavefront, sent out from a small dot, where the source was at that moment. Observer O sub one is ahead of it, and O sub two is behind. So, on which side are the wavefronts crowded? Yes, exactly, it is ahead of the source. The source chases its own waves, so the wavelength in front shrinks. Lambda prime, the Greek lambda with a small dash, is the new wavelength: v minus v sub s, over f. Here v is the speed of sound, and f is the frequency of the source. Shorter wavelength, so a higher pitch for O sub one. Okay? Behind the source, the waves spread out. Lambda prime equals v plus v sub s, over f, a longer wavelength, so O sub two hears a lower pitch. Right? Now keep the source still, and let the observer move towards it. The observer simply meets the waves more often, so it hears a higher frequency. Clear? Let me ask you this. The source moves towards you at some speed, or you move towards it at the same speed. Do you hear the same pitch both times? No, they are not the same, and this is a classic trap! One motion changes the wavelength, the other changes how often you meet the waves. So the two cases are not symmetric: always ask who is moving. Got it?

2:05The Doppler formula and its signs

Now, both effects in one formula. f prime, the frequency the observer hears, equals f times, v plus v sub o, over, v minus v sub s. The dash means the new value, as on lambda. Okay? Here v is the speed of sound relative to the air, the medium that carries the wave. With me? Then v sub o, the v with a small o below it, is the observer's speed, positive when it moves towards the source. Right? And v sub s is the source's speed, positive when it moves towards the observer. So, towards means positive, for both. Clear? Now, what if the source moves away from the observer? What sign does v sub s take then? It becomes negative, yes, exactly! Both speeds are relative to the air, and each is negative if it moves away. So for a receding source, the bottom becomes v plus that speed, and the pitch drops. Okay? Here is a quick check that toppers use. Moving towards each other, the pitch must go up; moving apart, it must go down. If your answer disagrees, a sign is wrong. Got it?

3:14Line of sight, wind and the passenger

Now a few rules that save marks. For motion at an angle, only the velocity components along the line joining them count. Right? Next, if wind of speed w blows from the source towards the observer, replace v with v plus w; the arrow means replace with. Okay? And if the source, the observer and the air are all at rest relative to one another, there is no shift at all. With me? Now, back to our puzzle. The driver moves along with the whistle. So what does the driver hear? The true frequency, that is correct! Anyone moving with the source hears no shift, so you and the driver were both right. Clear? For small speeds, there is a quick estimate. Capital delta, the small triangle, means the change in. So delta f over f is approximately v sub o plus v sub s, over v; the wavy equals sign means approximately. Okay? And light shifts too. Delta lambda over lambda is approximately u over c, where u is the speed of the source relative to the observer, along the line joining them, and c is the speed of light. For a receding source, the wavelength gets longer, which is called a red shift. Got it?

4:29Example 1: 500 Hz source approaching, then receding

Now our first example. The source has a frequency of five hundred hertz, the capital H and small z on screen. It moves at thirty metres per second; the small minus one on the s means per second. The speed of sound v is three hundred and thirty metres per second. What does a stationary observer hear as it approaches, and as it recedes? Pause the video and try it first. Done? Okay, let's check. The observer stands still, so v sub o is zero. Approaching, the source moves towards the observer, so v sub s is plus thirty. Okay? So f prime equals five hundred times three hundred and thirty, over three hundred and thirty minus thirty, which gives five hundred and fifty hertz. O sub one in the figure hears this higher pitch. With me? Now the source moves away. Going towards, the pitch went up by fifty. So going away, it comes down by fifty, to four hundred and fifty hertz, right? No, no, careful, that is exactly the trap here. Receding means moving away, so v sub s is now minus thirty. Got it? The bottom is now three hundred and sixty. So f prime equals five hundred times three hundred and thirty over three hundred and sixty, which gives about four hundred and fifty eight hertz, the lower pitch of O sub two. Not four hundred and fifty, since the rise and the drop are not equal. Clear?

6:00Example 2: train speed from a 6:5 drop in whistle pitch

Now, remember our train from the start? Let's find its speed. The pitch of its whistle drops in the ratio six is to five as it passes a stationary listener; the colon shows a ratio. The speed of sound is three hundred and thirty metres per second. Find the speed of the train. Pause the video and try it first. Done? Okay, let's check. Let u be the speed of the train. Approaching, f sub one equals f times v, over v minus u. Receding, the speed of the source is minus u, so f sub two equals f times v, over v plus u. Okay? But wait, we do not know f, the true frequency of the whistle. So how do we get rid of it? Take the ratio, yes, exactly! The f cancels, and f sub one over f sub two equals, v plus u, over, v minus u. Careful here: the pitch drops, so the higher frequency, f sub one, comes first and goes with the six. So the ratio is six over five. Right? Now cross multiply. Five v plus five u equals six v minus six u. Collect the terms, and eleven u equals v; the double arrow means which gives. With me? So u equals three hundred and thirty over eleven, which gives thirty metres per second. See, the drop in pitch alone gave us the speed of the train, as promised. Clear?

7:27Quick revision: formula, signs, source vs observer

So, quick revision. First, f prime equals f times, v plus v sub o, over, v minus v sub s, with towards positive and away negative. Second, the speeds are relative to the air, along the line joining source and observer. Third, a moving source and a moving observer are not symmetric. Now go and try the practice problems on this page. All the best!