Mixed Problems: Set 2 (Waves and Strings)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Two waves meet and interfere. The first has intensity capital I, and the second four times that, four capital I, okay? What is the largest intensity they can give? Five capital I, right? Hmm, are you sure? We will settle it in Example two. This is Set two of our mixed problems, waves and strings; questions like these come up often in JEE Main. So, let's begin.

0:31Tool 1: reading a travelling wave

So, tool one: reading a travelling wave. In the figure, y is the displacement at position x, and the wave moves at speed v. We write y equals capital A, times sine of k x minus omega t plus phi, where capital A is the amplitude, k the wave number and t the time. Omega, the Greek letter like a curly w, is the angular frequency, and phi, like a circle with a line through it, is the phase. Okay? Next, k is two pi over lambda; pi is the Greek letter on screen, and lambda, like an upside down y, is the wavelength. Omega is two pi f, with f the frequency. So the wave speed v is omega over k, which is f times lambda. Right? Now try one from our set. Here, y equals zero point zero two, times sine of two pi, times fifty t minus x over zero point four, in S I units. What is its speed? Okay, let's check. Two pi times fifty t is the omega t part, so f is fifty hertz, capital H z. Two pi x over zero point four is the k x part, so lambda is zero point four metres, the m on screen. With me? So v is f times lambda, fifty times zero point four; the cross means times. That gives twenty metres per second, written m s to the power minus one. If you got twenty, very good! Clear?

2:04Tool 2: particle speed is not wave speed

Now, tool two. Each bit of string only moves up and down, as the orange arrows marked particle velocity show. That velocity is the partial derivative of y with respect to t; the curly d means partial. For a wave moving towards plus x, as here, it equals minus v, times the slope of the string, the partial derivative of y with respect to x. Okay? Its largest value, the largest particle speed, is capital A times omega, which is not the same thing as the wave speed v. Got it? From our set: on a string, the wave speed is twenty metres per second, and the largest particle speed is two metres per second. Find the largest slope of the string. Just use the same relation. The largest slope, the modulus of the partial derivative of y with respect to x, with max written small below, is capital A omega over v; the two straight bars mean the modulus. So two over twenty, which is zero point one. That's it!

3:08Tool 3: standing waves on a string

Next, tool three: standing waves on a string fixed at both ends. In the figure, the letter N is a node, where the string never moves, and the letter A is an antinode, where it moves the most. Adjacent nodes are half a wavelength apart, lambda over two. Okay? From our set: adjacent nodes of a five hundred hertz standing wave are ten centimetres apart. Find the wave speed. Five hundred times ten centimetres, so fifty metres per second, right? No, no, careful, ten centimetres is only half a wavelength! So lambda is twenty centimetres, which is zero point two metres; the double arrow means so. Right? So v is f times lambda, five hundred times zero point two, which gives one hundred metres per second. With me? Now the modes of a string of length capital L, okay? f sub n, with the small n below, is n times v, over two capital L, which is n times f sub one, the fundamental. So all harmonics occur, as the three pictures show. Clear?

4:16Tool 4: tension sets the speed

Now, tool four. On a string, v is the square root of capital T over mu; the tick-shaped sign is the square root, the slash a fraction bar. This capital T is the tension, not the period, and mu, the Greek letter like a u with a tail, is the mass per unit length. So f sub n is n over two capital L, times root capital T over mu. Okay? For the same string and the same mode, n, capital L and mu stay fixed. So f is proportional to the root of the tension; the sign like an open alpha means proportional to. Right? From our set: the tension of a string is increased by forty four percent. By what percentage does its frequency rise? Forty four percent too, right? No, no, careful, the frequency goes as the square root! The tension becomes one point four four times, and the root of one point four four is one point two. So the frequency becomes one point two times, a twenty percent rise. Got it?

5:21Example 1: wave speed and fundamental of a 2 m string

Alright, Example one. Here, a two metre string of mass fifty grams is under a tension of eighty newtons; the g is grams, and capital N is newtons. Find the wave speed and the fundamental. Pause the video and try it first. Done? Okay, let's check. First, mu. Fifty over two gives twenty five, so mu is twenty five, right? No, no, wait, the grams must become kilograms first! Fifty grams is zero point zero five kilograms, so mu is that over two metres, which is zero point zero two five kilograms per metre, k g m to the power minus one. With me? Now v is the root of eighty over zero point zero two five, which is root three thousand two hundred. That is forty root two, about fifty six point six metres per second; the wavy equals sign means approximately. Right? And the fundamental, f sub one, is v over two capital L, with capital L two metres. So forty root two over four gives ten root two, about fourteen point one hertz. Clear?

6:32Example 2: max and min intensity when I and 4I interfere

Now a harder one, and it is our puzzle from the start. Example two: waves of intensities capital I and four capital I interfere. Find the largest and smallest resultant intensities. Pause the video and try it first. Done? Okay, let's check. For two waves of the same frequency with phase difference delta, the Greek letter like a curly d, the resultant intensity is I sub one plus I sub two, plus two root of I sub one times I sub two, times cosine delta. Here I sub one is capital I, and I sub two is four capital I, okay? So the largest is capital I plus four capital I, which is five capital I, right? No, no, careful, you forgot the last term! The largest comes when cosine delta is one: root I plus root four I, whole squared. Root four I is two root I, so that is three root I, squared, which is nine capital I, not five! So our puzzle is solved. With me? The smallest comes when cosine delta is minus one. Root I minus two root I is minus root I, and its square is capital I again. So the intensity swings between capital I and nine capital I, okay?

7:48Quick revision: wave speed, harmonics, tension, interference

So, quick revision. Read f and lambda off the equation, and v is f times lambda; the largest particle speed, capital A omega, is a different thing. Nodes are lambda over two apart, f sub n is n times f sub one, and on the same string, f goes as the root of the tension. In interference, the largest is root I sub one plus root I sub two, whole squared; the smallest has a minus sign instead. Now try the practice problem on this page. All the best!