Overview

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Take a wire whose fundamental is two hundred and fifty six hertz. We make its tension four times, and double its length. Does the note go up, go down, or stay the same? Keep your guess, and we will settle it. This chapter is about quick JEE Main questions on oscillations and waves. So, let's begin.

0:26How this chapter works

So, how does this chapter work? Oscillations and waves usually give two or three questions in a JEE Main shift. Okay? Usually one is on SHM, simple harmonic motion, or pendulums, and one or two are on strings, pipes, beats, or the Doppler effect. Right? The real questions from twenty twenty one to twenty twenty six are at the end of Chapters one to seven. Ours are original, about two minutes each. With me? There are three sets: SHM, waves and strings, and sound and Doppler. Let us pack the tools for each. Let's go.

1:04Set 1 tools: simple harmonic motion

Set one, simple harmonic motion. Small a is the acceleration, x is the displacement, and omega, the Greek letter like a curly w, the angular frequency. So a equals minus omega squared times x: the pull points back to the mean position, proportional to the displacement. That is the condition for SHM, okay? The motion is x equals capital A, times sine of omega t plus phi. Capital A is the amplitude, and phi, the circle with a line through it, is the initial phase. The period, capital T, is two pi over omega, with pi the Greek letter on screen. Right? Now let me ask you this: if we double the amplitude, does the period change? And what about the largest speed? The period does not change, because capital A is not in two pi over omega. But the largest speed doubles! v max, with the small subscript max, is capital A times omega, at the mean position, and a max is capital A times omega squared, at the extremes. With me? For a spring, capital T is two pi root m over k, with m the mass, k the spring constant, and the tick-shaped sign the square root. In parallel, the spring constants add, k one plus k two, with subscripts one and two. In series, the reciprocals add: one over k equals one over k one, plus one over k two. Clear? For a simple pendulum of length l, capital T is two pi times the square root of l over g, where g is the acceleration due to gravity, but only for small amplitude. And the mass of the bob does not appear. Got it?

2:48Set 2 tools: waves and strings

Set two, waves and strings. Wave speed v equals f times lambda, frequency times wavelength; lambda is the Greek letter like an upside down y. On a string, v is also the square root of capital T, the tension, over mu, the mass per unit length; mu is like a u with a tail, and the slash is a fraction bar. Careful, capital T here is the tension, not the period. With me? In the figure, a string of length capital L is fixed at both ends; the letter N marks a node, and the letter A an antinode. The modes are f sub n equals n times v, over two capital L, for n equal to one, two, three, and so on: all harmonics. Clear? Adjacent nodes are half a wavelength apart, lambda over two, so the node spacing gives the wavelength.

3:40Set 3 tools: sound and the Doppler effect

Got it? Set three, sound. With Laplace's correction, sound in a gas travels at the square root of gamma R T, over capital M, the molar mass. Gamma, the Greek letter like a y, is the ratio of the heat capacities, R the gas constant, and capital T now the temperature in kelvin. So v is proportional to root T, shown by the open alpha sign. Okay? On the right of the figure is a pipe open at both ends. Its modes are f sub n equals n times v, over two capital L, and all harmonics occur. Right? Now close one end, as on the left. The label there already says odd harmonics only, but why? The closed end must be a node, and the open end an antinode. So what lengths of the wave can fit in the pipe? Only odd numbers of quarter wavelengths, yes, exactly! Capital L can be lambda over four, as in the top picture, or three lambda over four, as in the bottom one, then five lambda over four, and so on. So the closed pipe gives two n minus one, in brackets, times v, over four capital L: one, three and five times the fundamental, and no second harmonic. Clear? Two sounds of nearly equal frequencies give beats, and the number per second is the modulus of f one minus f two; the two straight bars mean the modulus. With me? Last, the Doppler effect. The heard frequency, f prime, the f with a little dash, is f times a fraction: v plus v sub o, over v minus v sub s. Here v is the speed of sound, v sub o the speed of the observer, and v sub s the speed of the source. Okay? Now the signs, where students slip. v sub o is plus when the observer moves towards the source, and v sub s is plus when the source moves towards the observer; moving away, they are minus. And all speeds are relative to the air. Got it?

5:41Example 1: largest speed of a block released on a spring

Alright, Example one, from Set one. Here, a two kilogram block on a spring of two hundred newtons per metre is pushed in by zero point one metres and released. Find its largest speed. On screen, k g is kilograms, m is metres, and capital N, m to the power minus one is newtons per metre. Pause the video and try it first. Done? Okay, let's check. Released from rest, the block starts at an extreme, so the amplitude, capital A, is zero point one metres. Right? Next, from the spring period, omega is the square root of k over m. Two hundred over two is one hundred, and its root is ten radians per second, written r a d, s to the power minus one. Do not forget that root. With me? So v max is capital A times omega, zero point one times ten; the cross means times. That gives one metre per second, written m s to the power minus one. Clear?

6:43Example 2: wire with 4x tension and 2x length, new frequency

Now Example two, from Set two, a harder one, and it is our puzzle. Here, a wire's fundamental is two hundred and fifty six hertz; capital H z means hertz. Its tension is made four times, and its length doubled. Find the new fundamental. Pause the video and try it first. Done? Okay, let's check. Same wire, so mu stays the same. The fundamental is f equals one over two capital L, times the square root of capital T over mu, so f is proportional to root capital T, over capital L, okay? Now, many students say the tension is four times, so the frequency is four times, which is one thousand and twenty four hertz. Is that right? No, that is exactly the trap, so be careful here. The frequency goes as the square root of the tension, and the root of four is two. So the tension alone only doubles it. Right? But the length is doubled too, and f goes as one over the length, so that halves the frequency. With me? Together, f prime is two hundred and fifty six, times two over two, so still two hundred and fifty six hertz! Puzzle solved: the note stays the same. Got it?

8:02Quick revision: SHM period, string frequency, Doppler signs

So, quick revision. In SHM, the period does not depend on the amplitude, but v max and a max do. On a string, f goes as root T over L, and closed pipes give odd harmonics only. In the Doppler formula, speeds are relative to the air, and towards counts as plus. Now open Set one, and give each problem about two minutes. All the best!