Reflection, Echoes and Beats

Transcript

0:00Introduction

Hello everyone! Imagine you are driving straight towards a big wall, and you press the horn. You hear your horn and its echo together, and the sound throbs, louder and softer, many times every second. So how can the echo of your own horn come back at a different frequency? By the end, you will count those throbs exactly. Echoes come up often in JEE, so let's begin.

0:27The reflector: first an observer, then a source

First, look back at our Doppler picture. The circles are wavefronts from a source, capital S, the orange dot, moving right at v sub s, the speed of the source, shown by the orange arrow. Ahead, at O one, they crowd together: lambda prime, the Greek lambda with a little dash, is the new wavelength, v minus v sub s, over f, so the pitch is higher. Behind it, at O two, lambda prime is v plus v sub s, over f. Okay? In the Doppler formula, f prime, the f with a little dash, is the frequency heard: f, the true frequency, times, v plus v sub o, over v minus v sub s. Here v is the speed of sound relative to the air, v sub o is the speed of the observer towards the source, and v sub s is the speed of the source towards the observer, both measured relative to the air. If either one moves away, its speed goes in with a minus sign. Right? Now, a wall has no ears and no horn. So how can its echo come back at a different frequency? Here is the idea: the reflector plays two roles. In step one, it acts as an observer, and receives a frequency f sub one, the f with a small one below it. If the source comes towards it, like at O one in the picture, f sub one is higher than f. With me? In step two, it acts as a source, and sends out that same f sub one. So we use the Doppler formula twice. That is the whole trick. Clear?

1:56Car towards a wall: horn and listener both move

Here a car moves towards a wall at speed u, carrying both the horn and the listener, the driver, while the wall is at rest. Okay? In step one, the wall listens. So tell me, who is the source, who is the observer, and which one is moving? Yes, exactly, the car is the moving source and the wall the observer at rest. So v sub s equals the speed u, v sub o is zero, and f sub one equals f times v, over v minus u. Clear? In step two, the wall sends f sub one back. So the driver simply hears f sub one, right? No, no, wait, that is the JEE trap! The driver is moving too, so now the wall is a source at rest, and the driver is a moving observer. So the speed u goes on top: f double prime, the f with two little dashes, equals f sub one times, v plus u, over v. Got it? Put the two steps together, and v cancels. f double prime equals f times, v plus u, over v minus u. Write this down. With me?

3:03A surface moving towards a source at rest

Now flip it around: the source stays at rest, and a large surface comes towards it at speed u. In step one the surface listens, and it is the one moving. So where does the speed u go, on top or at the bottom? On top, that is correct, because the surface is an observer moving towards the source. So f sub one equals f times, v plus u, over v. Okay? In step two, the surface sends f sub one back while still moving, so it is a moving source: multiply by v, over v minus u. And look, we get f times, v plus u, over v minus u, exactly as for the car and the wall! Clear?

3:44Beats: the echo meets the direct sound

Now for the beats. The listener hears the echo and the direct sound together, and from the last chapter, the beats per second equal the difference of the two frequencies. Okay? But the driver moves along with his horn. So at what frequency does he hear his own horn? At the true frequency f, very good, because he moves with the source. So his horn at f and the echo at f double prime give him beats. Got it? So the beats are f double prime minus f. Here is a topper's shortcut: when the speed u is much smaller than v, shown by the double less-than sign, the beats are about two u over v, times f; the wavy equals sign means approximately. Clear? And one more case from the course: if a source sits between a listener and a wall, the direct sound and the echo may have different frequencies, and then the listener hears beats. Right?

4:37Example 1: 400 Hz car horn echo off a wall, and the beats

Now, remember our puzzle from the start? Let's solve it. So, a car approaches a wall at ten metres per second; the small minus one on the s means per second. Its horn is four hundred hertz, where hertz, the capital H and small z, is the unit of frequency, and v is three hundred and forty metres per second. Find the echo frequency the driver hears, and the beats. Pause the video and try it first. Done? Okay, let's check, with our car and wall formula. f double prime equals four hundred, times, three hundred and forty plus ten, over three hundred and forty minus ten. That is four hundred times three hundred and fifty, over three hundred and thirty. Okay? That gives about four hundred and twenty four hertz for the echo. Right? Now for the beats: the echo beats against which other sound, and at what frequency? Against his own horn, at the true four hundred hertz, because he moves with it. So the beats are four hundred and twenty four minus four hundred, about twenty four per second. Our puzzle is solved: that throbbing was about twenty four beats every second! And the topper's shortcut agrees. Two u over v, times f, is twenty over three hundred and forty, times four hundred, which gives about twenty three point five, also about twenty four.

6:00Example 2: sonar echo from an approaching submarine

Now a harder one. So, a ship's forty kilohertz sonar, where kilohertz means a thousand hertz, echoes off a submarine coming towards it at ten metres per second, and in water, v is one thousand five hundred metres per second. Find the echo frequency and the beat frequency. Pause the video and try it first. Done? Okay, let's check. The ship is at rest, and the submarine, the reflector, comes towards it. So it is our second case: f double prime equals f times, v plus u, over v minus u. Okay? So, with the speed in water, f double prime equals forty thousand, times one thousand five hundred and ten, over one thousand four hundred and ninety, which gives about forty point five four kilohertz. With me? Now for the beats: forty point five four minus forty is zero point five four kilohertz, so five hundred and forty hertz, right? Oh no, careful, that is the trap, because we rounded too early! Keep the digits: f double prime is about forty thousand five hundred and thirty seven hertz. Got it? Subtract the sonar's own forty thousand hertz, and the beat frequency is about five hundred and thirty seven hertz. When big numbers are close, keep every digit before subtracting. Clear?

7:20Quick revision: reflector rule, echo formula, beats

So, quick revision. First, a reflector is an observer first, then a source of what it received. Second, car to wall, or surface to a source at rest: f double prime equals f times, v plus u, over v minus u. Third, beats are the difference of the frequencies, about two u over v times f when u is much smaller than v. Now try the practice problem on this page. All the best!