Mixed Problems: Set 3 (Sound and Doppler)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Suppose a sound gets louder by thirty decibels. Thirty more decibels, so the sound is thirty times as intense, right? Hold on to that answer. Today is our third set of mixed problems, on sound and the Doppler effect, and problems like these come up often in JEE Main. So, let's begin.

0:25Idea 1: pipes, harmonics and overtones

Idea number one: organ pipes. The second overtone of an open pipe is six hundred hertz, the H z on screen. Find its fundamental. In the figure, the left pipes are closed at one end, at the thick black line, and the right pipes are open at both ends. The solid and dashed curves show how far the air moves, at two opposite moments, and each length, capital L, is written in terms of lambda, the wavelength; lambda is the Greek letter like an upside down y. With me? For a pipe open at both ends, f sub n, the f with a small n below it, equals n times v, over two capital L, where v is the speed of sound. Here n is one, two, three, and so on, so all harmonics, like f sub one and f sub two on the right. Clear? Now close one end. Then f equals, two n minus one in brackets, times v, over four capital L, okay? Two n minus one is always odd, so odd harmonics only, like f sub one and f sub three on the left. Got it? Next, overtones are the notes above the fundamental: the first overtone is the next note up, and the second is the one after that. Okay? So, for our open pipe, the second overtone is the second harmonic, and the fundamental is six hundred over two, which gives three hundred hertz, right? No, no, careful, that is the trap! The fundamental is the first harmonic, and it is not an overtone. So in an open pipe, the first overtone is the second harmonic, and the second overtone is the third harmonic. Right? So three times f sub one equals six hundred, and the double arrow on screen means which gives. So the fundamental, f sub one, equals six hundred over three, which gives two hundred hertz. Clear?

2:18Idea 2: decibels and intensity

Now, idea number two: our puzzle. The sound level rises by thirty decibels, written d B on screen. How much does the intensity change? Thirty times, as we guessed, or something else? Now, the sound level beta, the Greek letter like a tall B, equals ten times log to the base ten of, I over I nought, in decibels. Capital I is the intensity, and I nought, the I with a small zero, is ten to the power minus twelve watts per square metre, written W m to the power minus two. Okay? For two sounds, subtract the two levels. The I nought cancels, and beta sub two minus beta sub one equals ten log of, I sub two over I sub one. With me? Our rise is thirty, so thirty equals ten log of, I sub two over I sub one, and that log equals three. Right? So I sub two over I sub one is ten to the power three, which is one thousand. The sound becomes a thousand times as intense, not thirty times, and that solves our puzzle. Clear?

3:26Idea 3: the Doppler formula and its signs

Idea number three: the Doppler effect. Here, a listener moves away from a stationary source at half the speed of sound. What does he hear? The formula says f prime, the f with a small dash, the frequency heard, equals f times, v plus v sub o, over, v minus v sub s. Here f is the frequency of the source, and v is the speed of sound. Okay? Now, v sub o, the v with a small o below it, is the speed of the observer towards the source, and v sub s is the speed of the source towards the observer. Both are measured relative to the air, and moving away makes them negative: towards is plus, away is minus. Right? So, our listener moves away from the source at v over two. What sign does v sub o take? It is minus, yes, exactly! So v sub o is minus v over two, and the source is at rest, so v sub s is zero. Got it? So f prime equals f times, v minus v over two, over v. The v cancels, and f prime is f over two: he hears half the frequency. Clear?

4:39Example 1: source approaching at v/10, frequency ratio 10/9

Alright, example one. Here, a source approaches a stationary listener at one tenth of the speed of sound. Find the ratio of heard to emitted frequency. Pause the video and try it first. Done? Okay, let's check. Had the listener moved towards the source at v over ten, the ratio would be, v plus v over ten, over v, which is eleven over ten. Same speed now, so the same eleven over ten, right? No, no, careful, the two cases are not symmetric! In the figure, the orange dot, capital S, is the source, moving at v sub s, and the circles are wavefronts sent out from the small dots. Ahead, near O sub one, they crowd together: lambda prime, the new wavelength, is v minus v sub s, over f, a higher pitch. Behind, near O sub two, they spread out, a longer wavelength. Our listener is like O sub one: v sub s is plus v over ten, and v sub o is zero. Okay? So f prime over f equals v, over, v minus v over ten, which is v over, nine v over ten. Right? The v cancels, and the ratio is ten over nine, not eleven over ten. So here, the moving source raises the pitch a little more. Very good!

6:03Example 2: resonance tube, speed of sound and end correction

Now a harder one: the resonance tube, an air column closed at the water surface. With a five hundred hertz tuning fork, resonances occur at seventeen centimetres and fifty two centimetres. Find the speed of sound and the end correction. Pause the video and try it first. Done? Okay, let's check. The first resonance is a quarter wave, like the top left picture. So v equals four times f times l sub one, the first length: four times five hundred times zero point one seven, which gives three hundred and forty metres per second, right? Oh no, careful, that forgets the end correction! The antinode lies a little beyond the open top, and that small extra length is e, the end correction. So l sub one plus e equals lambda over four, and l sub two plus e equals three lambda over four, like the two left pictures. With me? We do not know e, so subtract the two, and e cancels. l sub two minus l sub one equals lambda over two, and fifty two minus seventeen gives thirty five centimetres. Okay? So lambda is seventy centimetres, which is zero point seven metres. Then v equals f times lambda: five hundred times zero point seven zero gives three hundred and fifty metres per second, written m s to the power minus one. Right? Now the end correction. The end correction e equals, l sub two minus three l sub one, over two. Three times seventeen is fifty one, and fifty two minus fifty one, over two, gives zero point five centimetres. So the speed is three hundred and fifty, not three hundred and forty. Very good!

7:51Quick revision: overtones, dB, Doppler signs, resonance tube

So, quick revision. First, an open pipe's second overtone is its third harmonic, and a closed pipe gives odd harmonics only. Second, thirty decibels more means a thousand times the intensity. Third, in the Doppler formula, towards is plus, away is minus, and source and listener are not symmetric. Fourth, in the resonance tube, v equals two f times, l sub two minus l sub one, and e equals, l sub two minus three l sub one, over two. Now try the practice problem on this page. All the best!