Normal Modes of Strings and Rods

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a string fixed at both ends. It resonates at four hundred and twenty hertz, and its very next resonance is at four hundred and ninety hertz. So what is its lowest frequency, the fundamental? Is it four hundred and twenty? Keep your answer in mind; we will settle it today. This comes up often in JEE, so let's begin.

0:27String fixed at both ends

So, here is a string of length capital L, tied firmly at both ends. The figure shows its first three normal modes; the letter N marks a node, and the letter A marks an antinode. Now tell me, what must the string do at a tied end? It cannot move at all, yes, exactly! So both fixed ends are nodes, points of zero amplitude. Okay? And adjacent nodes are half a wavelength apart, remember? So the length must hold a whole number of half wavelengths. Capital L equals n times lambda over two, where lambda, the Greek letter, is the wavelength. And n is one, two, three, and so on, as the three dots say. Right? Now, the speed on a string is the square root of capital T over mu, the tension over the mass per unit length; mu is the Greek letter like a u with a tail. Speed over wavelength gives the frequency. So f sub n equals n over two capital L, times the square root of capital T over mu. The small n below marks the n th mode, and the tick-shaped sign is the square root. Clear? Every n is allowed, so all harmonics occur; n equal to one is the fundamental, the first harmonic. Now here is a classic trap. Is the second harmonic the same as the second overtone? No, careful, they are not the same! The overtones are counted from the first mode above the fundamental. So the n th harmonic is the overtone numbered n minus one, in the brackets on screen; the second harmonic is the first overtone. Got it? Now count in the figure. The n th mode has n loops, and n plus one nodes, counting both ends. In the bottom picture, three loops and four nodes. With me?

2:12Successive resonances, and a free end

Now, from that formula, every mode is a whole number times the fundamental, right? So tell me, what is the gap between two successive resonant frequencies? It is exactly the fundamental, yes! For a string with both ends fixed, f sub n plus one, the next mode, minus f sub n, equals f sub one, and our puzzle needs exactly this. Okay? Now a different string. One end is fixed, and the other end has a light ring that slides on a smooth rod, as the brackets on screen say. So that end is free to move. Is it a node, or an antinode? It is an antinode, yes, exactly! The fixed end is a node, the free end an antinode, and a node to the next antinode is a quarter wavelength. So capital L equals two n minus one, in brackets, times lambda over four, an odd number of quarter wavelengths. And f equals two n minus one, times v over four capital L, where v is the wave speed. Clear? Now, two n minus one gives only one, three, five, and so on. So only odd harmonics occur: f sub one, three f sub one, five f sub one. The even ones are simply missing. So here, neighbouring resonances differ by two f sub one, not by f sub one. Got it?

3:31Sonometer, energy and rods

Next, the sonometer. Here we change the vibrating length of a wire, small l, until it resonates with a tuning fork. The resonating length satisfies f equals one over two small l, times the square root of capital T over mu. With me? Now let me ask you this. With the same tuning fork, f stays fixed. So how must small l change with the tension? Small l is proportional to the square root of capital T; the open loop sign means proportional to. So the length goes as the root of the tension, not the tension itself. Careful there, okay? Now the energy of a normal mode. For the n th mode with amplitude small a, the amplitude at an antinode, capital E equals one quarter, times m, times omega sub n squared, times small a squared. Here m is the mass of the whole string, and omega sub n, the Greek letter omega, is the angular frequency of that mode. Right? Finally, longitudinal waves in a rod travel with speed v equals the square root of capital Y over rho; the slash is a fraction bar. Capital Y is Young's modulus, and rho, the Greek letter like a curly p, is the density. Clear? And in a rod, free ends are antinodes, and clamped points are nodes.

4:50Example 1: first three frequencies of a 1 m string at 100 N

Got it? Example one: a string one metre long, with mu equal to zero point zero one kilograms per metre, is under a tension of one hundred newtons. Find the first three natural frequencies. On screen, k g m to the minus one is kilograms per metre, and capital N is newtons. Pause the video and try it first. Done? Okay, let's check. First, the speed. Capital T over mu is one hundred over zero point zero one, which is ten to the power four; the small four up top is the power. Its square root is one hundred, so v is one hundred metres per second, written m s to the minus one. With me? Now the fundamental. Both ends are fixed, so which formula do we use, and what do you get? f sub one equals v over two capital L, yes! That is one hundred over two times one, and the cross means times. So f sub one is fifty hertz; capital H z means hertz. Okay? And all harmonics occur, so the next two are just two times and three times the fundamental. f sub two is one hundred hertz, and f sub three is one hundred and fifty hertz. Right?

6:04Example 2: wave speed from resonances at 420 and 490 Hz

Now a slightly harder one, and it is our puzzle from the start. Example two: a string seventy centimetres long, fixed at both ends, has two successive resonances at four hundred and twenty hertz and four hundred and ninety hertz. Find the wave speed. On screen, c m means centimetres. Pause the video and try it first. Done? Okay, let's check. Many students take four hundred and twenty hertz as the fundamental and go straight to the speed. Is that right? No, careful, that is the trap! Nobody said four hundred and twenty is the lowest mode. Two successive resonances differ by the fundamental, remember? So f sub one is four hundred and ninety minus four hundred and twenty, which is seventy hertz. Clear? Let us quickly check this. Four hundred and twenty is six times seventy, and four hundred and ninety is seven times seventy. So these are the sixth and seventh harmonics, neighbours, exactly as they should be. And that settles our puzzle: the fundamental is seventy hertz, not four hundred and twenty. Got it? Now the speed. From f sub one equals v over two capital L, we get v equals two capital L times f sub one. Seventy centimetres is zero point seven metres. So v equals two times zero point seven times seventy, which gives ninety eight metres per second.

7:28Quick revision: harmonics, successive resonances, sonometer

Okay? So, quick revision. For a string fixed at both ends, f sub n equals n over two capital L, times the square root of capital T over mu, and all harmonics occur. Here, two successive resonances differ by the fundamental, f sub one. With one end free, only odd harmonics occur, and in a sonometer, small l is proportional to the square root of capital T, the tension. Now go and try the practice problems on this page. All the best!