Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take two tuning forks: one of three hundred hertz, and one we do not know. Together, they throb, loud and soft, three times a second. Now we stick a little wax on the unknown fork, and the throbbing gets faster. So, is the unknown fork above three hundred hertz, or below? Hold your answer; our last example settles it. This is part A of our unit problem bank, at JEE Main level: three quick tools, then two full problems. So, let's begin.

0:35Tool 1: omega and amplitude from v max and a max

So, tool one. In SHM, simple harmonic motion, v max, the v with a small max below it, is the largest speed: zero point five metres per second, written m s to the power minus one. And a max, the largest acceleration, is two point five metres per second squared, written m s to the power minus two. Find omega, the Greek letter like a rounded w, the angular frequency, and capital A, the amplitude, okay? The speed is largest at the mean position, where v max equals capital A times omega. The acceleration is largest at the extremes, where a max equals capital A times omega squared. Clear? Now, we have two equations and two unknowns. So tell me, what is the fastest way to get omega? Divide them, yes, exactly! Dividing a max by v max gives capital A omega squared, over capital A omega. The amplitude and one omega cancel, leaving just omega. With me? So omega is two point five over zero point five, which gives five radians per second, written rad s to the power minus one. Right? Then the amplitude is v max over omega: zero point five over five gives zero point one metre, where small m is metres. Divide first, and the rest follows. Got it?

1:53Tool 2: a pendulum in a lift

Now tool two. Here, a pendulum of period capital T hangs in a lift accelerating downwards at g over two, half the acceleration due to gravity. Find its new period, okay? For a small amplitude, a simple pendulum has T equals two pi times the square root of l over g, with small l the length. The sign like a stretched alpha means proportional to, so T is proportional to one over root g. Clear? In an accelerating lift, we replace g by g effective, the g with a small e f f below it. When the lift accelerates downwards at small a, g effective equals g minus small a. Right? Here small a is g over two, so g effective is g minus g over two, which is just g over two. With me? So tell me: g effective is halved, so the period doubles, right? No, no, careful, that is the trap! T is proportional to one over root g, not one over g. So T prime, the new period, the T with a small dash, is two pi times the root of, l over, g over two, which is root two times T, okay? That is about one point four one times T; the wavy equals sign means approximately. Slower, but not twice as slow. Got it?

3:12Tool 3: close one end of an open pipe

Now tool three, on sound. An open pipe has a fundamental of three hundred hertz, the H z on screen. One end is then closed. Find the new fundamental. In the figure, the left pipes are closed at one end, and the right ones are open at both ends, okay? In an open pipe, both ends are displacement antinodes, so the simplest fit is half a wavelength, lambda over two, the top right picture; lambda, the Greek letter like an upside down y, is the wavelength. So f sub one, the fundamental, equals v over two capital L, with v the speed of sound and capital L the length. Right? Now we close one end. Does the fundamental go up, or down? It goes down, that is correct! The closed end is a displacement node, so the simplest fit is just a quarter of a wave, the top left picture. So now f sub one equals v over four capital L, clear? The length and the speed of sound have not changed. So the closed fundamental, v over four capital L, is exactly half of v over two capital L, with me? So the new fundamental is three hundred over two, which gives one hundred and fifty hertz. And as the figure says, a closed pipe gives odd harmonics only. Got it?

4:30Example 1: f, wavelength and speed from a wave equation

Now, example one. y equals three times the sine of, two pi times, t over zero point zero four, minus x over forty, in big brackets. Here y and x are in centimetres, c m, and t is in seconds. Find f, the frequency, lambda, the wavelength, and v, the wave speed. Pause the video and try it first. Done? Okay, let's check. Open the big brackets: the term with t is two pi over zero point zero four, times t. That number is omega, and omega equals two pi f. So f equals one over zero point zero four, which gives twenty five hertz. Right? Now the term with x is two pi over forty, times x, and that number is k, the wave number, which equals two pi over lambda. Since x is in centimetres, lambda is forty centimetres. With me? Now v equals f times lambda, and twenty five times forty gives one thousand. So the wave speed is one thousand metres per second, right? Oh no, careful, look at the units! Lambda was in centimetres, so f times lambda gives one thousand centimetres per second, written c m s to the power minus one. Clear? And one thousand centimetres is ten metres, so v is ten metres per second. Always check the units first. Okay?

5:55Example 2: waxed fork gives more beats, 297 or 303 Hz?

And now, our puzzle from the start, in the bank's words: a fork gives three beats per second with a three hundred hertz fork, and waxing it increases the beat rate. Find its frequency. Pause the video and try it first. Done? Okay, let's check. Beats per second equal the modulus of, f sub one minus f sub two; the two straight bars mean the positive value. So the unknown f is three hundred, plus or minus three; the plus or minus sign means both. That is three hundred and three hertz, or two hundred and ninety seven hertz. Okay? Many students say this: more beats, so the higher one, three hundred and three hertz, right? No, no, wait, that is the trap! Loading a prong with a little wax lowers the frequency slightly; a prong is one of the fork's two arms. So test both. If the fork were three hundred and three, lowering it moves it towards three hundred, so the difference shrinks and the beats become fewer. With me? But from two hundred and ninety seven, lowering it moves it away from three hundred. The difference grows, and the beats become more, exactly as the question says. Clear? So the fork's frequency is two hundred and ninety seven hertz. That settles our puzzle: the unknown fork was below three hundred. Change one source, test both candidates. Very good!

7:19Quick revision: omega, g eff, closed pipe, units, beats

So, quick revision. In SHM, omega equals a max over v max. For a pendulum, use g effective, g minus small a in a lift accelerating down, and remember the square root. Next, a closed pipe's v over four capital L is half the open pipe's v over two capital L; and in waves, watch the units. And for beats, f equals f sub zero, the known frequency, plus or minus n, the beats per second. Wax lowers f, so test both candidates. Now try the practice problem on this page. All the best!