Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Tie one end of a long rope to a wall, and flick the other end. Look, a hump runs all along the rope, right up to the wall. So, did any piece of the rope travel to the wall along with it? Hold that thought; we will settle it in Example one. Today we work through five concept checks from this unit's problem bank. Ideas like these are tested in JEE, often as statements. So, let's begin.

0:32Check 1: the pendulum at large swings

So, check one. For a simple pendulum of small amplitude, capital T, the period, equals two pi times the square root of l over g. Pi is the Greek letter, l is the length of the string, and g is the acceleration due to gravity. This is for small amplitude, okay? Now the check. Pull the bob far out and let it swing wide. Is the period still independent of amplitude? You know the rule: the period of simple harmonic motion, SHM, does not depend on the amplitude. So, yes, correct? Oh no, careful here, that rule is only for SHM, and a wide swing is not SHM at all. The restoring force on the bob is m g sine theta, where m is the mass of the bob, and theta, this Greek letter, is the angle of the string from the vertical. It is not m g theta. With me? SHM needs a restoring force proportional to the displacement; this sign, like an open alpha, means proportional to. And sine theta is about equal to theta only for small angles; the wavy equals sign means approximately. So only for small swings is the motion simple harmonic. Clear? For large swings, m g sine theta is no longer proportional to theta, so the motion is not simple harmonic, and the formula does not hold. So the answer is no: the period grows with amplitude. Got it?

1:56Check 2: moving at right angles to the source

Now check two, the Doppler effect. f prime, the f with a small dash, is the frequency the listener hears, and f is the frequency of the source. v sub o, the v with a small o below, is the speed of the observer, and v sub s is the speed of the source. So f prime equals f, times v plus v sub o, over v minus v sub s. Okay? Here v is the speed of sound relative to the air. In the figure, the source, the red dot marked capital S, moves to the right, along the red arrow marked v sub s, so its wavefronts crowd together ahead of it, near O one, and spread out behind it, near O two. So the wavelength, lambda prime in the labels, is shorter ahead and longer behind. Right? Now, the signs matter a lot. v sub o and v sub s are both measured relative to the air. Each is positive when moving towards the other, and negative when moving away. With me? Now the check. The listener moves at right angles to the line joining him to the source. He is moving, so surely the pitch changes, right? Is there a Doppler shift? No, his motion gives no shift. Only the velocity components along the line joining the source and the listener count, the line of sight. Moving at right angles, his velocity has no component along that line, so v sub o is zero. Clear?

3:21Check 3: when beats are not heard

Alright, check three, on beats. Take two sounds of nearly equal frequencies, f one and f two, with the small one and two written below. The number of times their loudness rises and falls each second is mod of f one minus f two. These two vertical bars mean mod, the modulus, so the difference is taken as positive. Okay? Now the check. When f one and f two differ by more than about ten hertz, we do not hear beats. H z, short for hertz, means per second. Has the formula stopped working? No, the formula is fine; the problem is the ear. The ear cannot follow such rapid rises and falls of loudness. They come too fast. Right? So instead of separate beats, it hears a rough tone. Got it?

4:09Example 1: does a travelling wave transport matter?

Now Example one, and it is our rope puzzle. Does a travelling wave transport matter? Pause the video, and try it first. Done? Okay, let's check. The figure shows the wave at one instant: y, the displacement of each particle, against x, its position along the string. The wave moves to the right with speed v, the black arrow; lambda, the Greek letter, is the wavelength, and capital A is the amplitude. Now the small red arrows, labelled particle velocity, show how each particle moves, and they point only up or down, never along the wave. Right? So, back to our rope. The hump travelled all the way to the wall, so the rope under the hump must have travelled with it, correct? No, no, wait, look at those red arrows again. Each particle only oscillates about its own mean position, up, down and back again. It never travels along with the wave. With me? So the answer is no. The wave carries energy and momentum, but not matter. On our rope, only the disturbance reached the wall; every piece of rope stayed in its place, apart from its small up and down motion. That settles our puzzle. Clear?

5:25Example 2: are pipe displacement nodes also pressure nodes?

Now Example two, a little harder. Do the displacement nodes of a sound wave in a pipe coincide with its pressure nodes? Here is a hint: think about the air right at a closed end. Pause the video, and try it first. Done? Okay, let's check. The figure shows how far the air moves along each pipe: on the left, pipes closed at one end, and on the right, pipes open at both ends. At a closed end, the air cannot move, so it is a displacement node. Okay? At an open end, the air moves the most, so it is a displacement antinode. The labels give each pipe's length, capital L, in terms of lambda, and its frequency: odd harmonics only for the closed pipe, all harmonics for the open one. Right? Now the tempting answer. At the closed end, the air does not move at all. So it must be a pressure node as well, correct? Oh no, careful here, we need the pressure, not the displacement. The displacement wave is s equals s nought, times sine of omega t minus k x. s nought, with the small zero below, is the displacement amplitude, k is two pi over lambda, and omega, the Greek letter, is the angular frequency. Okay? And the pressure wave is delta p equals p nought, times cosine of the same omega t minus k x. Delta, the little triangle, means change, so delta p is the change in pressure, and p nought is the pressure amplitude. Sine for one, cosine for the other: the pressure is a quarter wave out of step with the displacement. With me? So the answer is no. Displacement nodes are pressure antinodes. At the closed end, the air does not move, but its pressure changes the most. Right? And it works the other way round too. At an open end, a displacement antinode, the pressure does not change: it is a pressure node. So the two sets of nodes never coincide.

7:24Quick revision: large-swing pendulum, Doppler, beats, nodes

Got it? So, quick revision. The pendulum formula holds only for small swings; for large swings, the period grows. In the Doppler effect, only the components along the line of sight count, and beyond a difference of about ten hertz, the ear hears a rough tone. Waves carry energy and momentum, not matter, and displacement nodes are pressure antinodes. Now try the practice question on this page. All the best!