Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Sound a tuning fork together with a three hundred hertz fork, and you hear three beats every second. Now put a little wax on the first fork, and the beats get faster. So, was the first fork two hundred and ninety seven hertz, or three hundred and three hertz? Hmm, it could be either, right? Keep your guess. First, we will see how to use this unit's problem bank, and then we will settle this puzzle with a problem from it. Puzzles like this are a common JEE pattern. So, let's begin.

0:36What is in this chapter

So, first, let us see what this chapter is. It is the problem bank for the whole unit on oscillations and waves: mixed solved problems, in three parts. Okay? Part A has six solved problems, at JEE Main level. Part B has five longer problems, at JEE Advanced level. With me? And Part C has five concept checks, on the ideas that are often tested in statement form. And each part ends with one practice problem on the page, for you to solve. Got it?

1:08Part A: a 30-minute test

Now, when should you attempt Part A of this chapter? On the very first day of the unit, just to have a look? No, not on the first day: the course says to attempt it only after Chapters one to eight, okay? That is because its problems are mixed: pendulums, springs, SHM, that is simple harmonic motion, pipes, waves and beats, all together. So finish every chapter first. Right? Then attempt it as a real thirty minute test: solve each problem yourself first, and only then check the solution. With me?

1:44Part C: two statement checks

Now, a taste of Part C, with two checks just as the course asks them. Here is the first one: is the period of a simple pendulum independent of amplitude, even for large swings? The answer is no. The restoring force is m g sine theta, not m g theta. Here m is the mass of the bob, g the acceleration due to gravity, and theta, the Greek letter, is the angle of the string from the vertical. Okay? Only for small swings is sine theta about equal to theta; the wavy equals sign means approximately. Then the restoring force is proportional to the displacement, the motion is SHM, and capital T, the period, equals two pi root of l over g, with small l the length. For large swings, the period grows with amplitude. Statements like this come up often, so read them carefully. Right? Now the second check. The listener moves at right angles to the line joining him to the source. Is there any Doppler shift? There is no shift at all, because only the component of the velocity along that line counts, and here that component is zero. Clear? And Part B has a twin of this one: a source moving on a circle, with the listener at its centre. The source's velocity is always perpendicular to the line joining it to the listener, so again, no shift is heard. Got it?

3:07Example 1: pendulum in a lift accelerating down at g/2

Now Example one, from Part A, an easy one. Here, a pendulum of period capital T hangs in a lift, and the lift is accelerating downwards at g over two, half of g. Find its new period. Pause the video, and try it first. Done? Okay, let's check. Inside the lift, the pendulum feels an effective gravity, g effective, written g with the small subscript e f f. Many students say the lift is accelerating, so just add: g plus g over two. Is that right? No, no, careful, the direction matters! For a lift accelerating down at small a, g effective is g minus small a, okay? It is g plus small a only when the lift accelerates up, as in the bracket. Clear? Here, small a is g over two. So g effective equals g minus g over two, which is g over two. Inside this lift, gravity feels only half as strong. Right? Now, T prime, the T with a little dash, is the new period. T prime equals two pi times the root of small l over g effective, where small l is the length of the pendulum. With g over two in the bottom, that becomes two pi root of two l over g. With me? Take the two out of the root as root two; the cross means times. So T prime is root two, times two pi root of l over g. And two pi root of l over g is just capital T, the old period. So T prime equals root two times T: a longer period, because weaker gravity gives slower swings. Got it?

4:42Example 2: waxed fork, 3 beats with 300 Hz: 297 or 303?

Now Example two, Part A's medium problem, and it is our puzzle from the start. Here, a fork gives three beats per second with a three hundred hertz fork; the capital H and small z mean hertz. Waxing it increases the beat rate. Find its frequency. Pause the video, and try it first. Done? Okay, let's check. The beats per second equal the difference of the two frequencies: the modulus of f sub one minus f sub two, where the vertical bars mean modulus. So f, the fork's frequency, is three hundred plus or minus three, where the plus sign over the minus sign means plus or minus. That gives three hundred and three, or two hundred and ninety seven hertz. Okay? Now, which one is it? Tell me, when we put wax on a fork's prongs, does its frequency go up, or down? Yes, it goes down: loading a prong with wax lowers the fork's frequency. Remember this one, right? Now, many students see the word increases, and pick the higher value, three hundred and three hertz. Is that right? No, no, wait, let us test it properly. If f were three hundred and three, the wax would pull it down, nearer to three hundred. The difference would shrink, so the beats would fall. But the question says they increase. With me? Now test two hundred and ninety seven. The wax pulls it further below three hundred, so the difference grows, and the beats rise, exactly what the question says. Clear? So the fork's frequency is two hundred and ninety seven hertz, and our puzzle is settled. If that was your guess, very good! The method is always the same: write both answers, then see which way the change moves the beats. Got it?

6:31Quick revision: g eff in a lift, beats and the wax rule

So, quick revision. Part A is a thirty minute test after Chapters one to eight; Part B is at JEE Advanced level; and Part C has the concept checks. The pendulum formula, two pi root of l over g, is for small swings only, and in a lift accelerating down, g effective is g minus small a, while accelerating up, it is g plus small a, okay? And beats are the difference of the two frequencies; wax lowers a fork's frequency, so test both answers. Now set a thirty minute clock, open Part A, and begin. All the best!