Speed, Energy and Power of Waves on a String

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a long rope hanging from the ceiling, and give its bottom end a quick flick. Now a little pulse runs up the rope, towards the ceiling. Does it climb at a steady speed, or does it speed up, or slow down? Keep your answer in mind; we will settle it soon. In this lesson, we find how fast a wave runs on a string, and how much energy and power it carries. This comes up often in JEE, so let's begin.

0:30Wave speed on a string

So, here is the key result. The speed of a wave on a string, small v, equals the square root of T over mu. The tick-shaped sign is the square root sign, and mu is the Greek letter that looks like a u with a tail. Okay? Capital T is the tension, in newtons, the capital N in the bracket. And mu is the mass per unit length; on screen, k g, then m to the power minus one, means kilograms per metre. Right? Now let me ask you this. If we double the tension in the string, does the wave speed also double? If you said yes, careful, that is the trap! The tension sits under a square root, so when T becomes two times, v becomes only root two times, not two times. Clear? Now, for a wire, we usually know the density instead. Rho, the Greek letter that looks like a p, is the density, and small a is the area of cross section. Then mu equals rho times a, the mass of one metre of wire, so v equals the square root of T, over rho times a. Got it?

1:35A rope hanging from the ceiling

Now back to our hanging rope. Let x be the height of a point above the lower end. The tension there holds up all the rope below it, a piece of mass mu times x, and weight mu times x times g, with g the acceleration due to gravity. So T equals mu times g times x. Okay? So, as the pulse climbs, x, the height, grows, and so does the tension. Now tell me: going up the rope, does the pulse speed up, or slow down? It speeds up, yes, exactly! Put this tension into our formula: the square root of mu times g times x, over mu. The mu cancels, so v equals the square root of g times x. Higher up, x is bigger, so the pulse runs faster. That settles our puzzle. Right? Now, let capital L be the length of the rope, and we want the time for the pulse to reach the top. Can we simply take the length over the speed? No, we cannot, and that is the trap, because the speed keeps changing on the way up. So we add up tiny bits of time: each little length d x takes d x over the square root of g times x. The tall curly S is the integral sign, adding from zero, at the bottom, up to capital L, at the top. The result is t equals two times the square root of L over g. With me? And one more case. If a block of mass m hangs at the bottom, the tension must hold it up too. So T equals the bracket m plus mu times x, all times g. Clear?

3:11Frequency stays, wavelength changes

Alright, next idea. Suppose a wave passes from one string into another, where it travels more slowly. What stays the same, the frequency or the wavelength? The frequency, exactly! The frequency f is fixed by the source that shakes the string. It does not change from one medium to another. Okay? So the wavelength must change. Since v equals f times lambda, where lambda, the Greek letter, is the wavelength, lambda is proportional to v; the curly sign means proportional to. In the slower string, the wavelength is shorter. Right?

3:49Energy and power carried by the wave

Now, a wave also carries energy. The energy per unit length is one half times mu times omega squared times capital A squared; the small two up top means squared. Omega, the Greek letter that looks like a curly w, is the angular frequency, and capital A is the amplitude. Okay? And the power, capital P, is the energy carried along each second. P equals one half times mu times omega squared times A squared, times v. And why is the v there? Because energy per metre, times metres per second, gives energy per second. With me? Quick question now. Keep everything else the same, and double the amplitude. What happens to the power? If you said it doubles, careful, that is the trap! The power is proportional to A squared, so double the amplitude, and the power becomes four times. Got it?

4:46Example 1: wave speed on a string, T 90 N and mu 0.1 kg/m

Now, an example. Example one: find the wave speed on a string of mass per unit length zero point one kilograms per metre, under a tension of ninety newtons. Pause the video and try it first. Done? Okay, let's check. The speed is the square root of T over mu, so the square root of ninety over zero point one. Okay? Now, ninety over zero point one. Many students quickly write nine here. Is that right? No, that is the trap, so be careful! Ninety over zero point one is the same as ninety times ten, which gives nine hundred, not nine. Right? So v equals the square root of nine hundred, which is thirty metres per second. On screen, m, then s to the power minus one, means metres per second. Clear?

5:38Example 2: power of a 50 Hz wave on a string under 80 N

Now a slightly harder one. Example two: a string with mu equal to zero point zero five kilograms per metre, under eighty newtons, carries a fifty hertz wave of amplitude one centimetre. Find the power transmitted. On screen, capital H z is hertz, and c m is centimetres. Pause the video and try it first. Done? Okay, let's check. First, the speed: eighty over zero point zero five gives one thousand six hundred, and its square root is forty, so v is forty metres per second. Okay? Now, the power formula has omega squared in it, and we are given the frequency, fifty hertz. Do we put fifty straight in for omega? No, careful, that is a very common slip! Omega is the angular frequency, two pi times f. Two pi times fifty gives one hundred pi radians per second, the r a d, s to the power minus one, on screen. Right? And the amplitude must be in metres. One centimetre is ten to the power minus two metres, so the amplitude squared is ten to the power minus four square metres. Clear? Now put everything in. P equals one half, times zero point zero five, times one hundred pi whole squared, times ten to the power minus four, times forty. Okay? Look for the nice pairs. One half times zero point zero five times forty is exactly one. And one hundred pi whole squared, times ten to the power minus four, is just pi squared. With me? So P equals pi squared, about nine point eight seven watts; the wavy equals sign means approximately, and capital W is for watts. That was quick, once the units were right. Got it?

7:30Quick revision: string speed, hanging rope, wave power

So, quick revision. On a string, the speed is the square root of T over mu; on a hanging rope, it is the square root of g times x. The frequency is fixed by the source, and the wavelength is proportional to the speed. And the power is one half times mu times omega squared times A squared times v, with omega equal to two pi times f, and the amplitude in metres. Now go and try the practice problems on this page. All the best!