The Simple Pendulum

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take two pendulums of the same length, with bobs of the same size, but one bob is heavy and the other is light. Let both go from a few degrees. Which one swings faster? We will settle it soon. Pendulum questions come up often in JEE, under experimental skills. So, let's begin.

0:24Finding g from the period

So, look at the pendulum on the left of the picture, and the graphs beside it come later, okay? The period, capital T, equals two pi times the square root of L over g. Pi is the Greek letter, L is the length marked in the drawing, capital A is the amplitude, and g is the acceleration due to gravity. So what do we do to this formula to find g? Yes, exactly, we square both sides and make g the subject. So g equals four pi squared times L, over capital T squared. So all we have to measure is the length L and the period capital T, right? Now look at this L carefully. You hold a scale against the thread and read its length. Is that length your L? No, it is not, and that is a real trap. L is the effective length, from the point of suspension to the centre of the bob. So it is the thread, plus the hook, plus the radius of the bob. The bob in our picture has no hook, so its note leaves that part out. Clear?

1:28Doing the experiment properly

Now, back to our puzzle. Look at this formula for the period, and search for the mass of the bob. So which bob swings faster? The answer is that neither swings faster, and that surprises many students. There is no mass in the formula. And for small swings the period does not depend on the amplitude either, so we keep it to a few degrees, okay? One oscillation would be so much quicker to time, right? So why do we time twenty or more? See, the error of the stopwatch and of your reaction is fixed, so it is a much smaller fraction of a long time. Delta, the little triangle, means the error in a quantity, and n is the number of oscillations. So delta capital T equals delta t over n, with small t the total time. With me? Here is one more practical point. Where should you start counting, at the extreme position, or at the mean position? It is the mean position, because the bob moves fastest there, so the instant it passes is sharply defined. At the extremes it is almost at rest for an appreciable time.

2:39Graphs and the seconds pendulum

Right? Now let us draw a graph. To get a straight line, should you plot L against capital T, or L against capital T squared? That is correct, it is capital T squared. L against capital T squared is a straight line through the origin, and its slope is g over four pi squared. Clear? If you plot L against capital T itself, you get a parabola instead. Remember that shape, okay? Now be careful here, because marks are often lost. If a question plots capital T squared against L, the axes are swapped, and the slope turns upside down, to four pi squared over g, right? Now the seconds pendulum, and its name is a small trap. Its period is two seconds, not one second. Its length is about ninety nine point three centimetres, and the wavy equals sign means approximately. Got it?

3:33The error in g

Now let us find the error in g from this formula. Which error counts double, the one in L or the one in capital T? Yes, exactly, the one in capital T, because it is squared in the formula. So delta g over g equals delta L over L, plus two times delta capital T over capital T, okay? So careful timing matters. With me?

3:59Why the swings slowly die out

Now, air resistance and friction at the support take energy from a pendulum, so its swings slowly die out. Capital A equals capital A nought times e to the power minus b times t, over two m. Capital A nought, with the small zero, is the starting amplitude, t is time, m is the mass, and b measures the damping. Okay? The energy, capital E, is one half times m times omega squared, times capital A squared. Omega is a Greek letter, and omega squared equals g over L, right? Now, when we square capital A, what happens to the power of e? That is right, the power doubles. So the energy is proportional to capital A squared, which is capital A nought squared times e to the power minus b times t over m. The curly symbol means proportional to. Clear? And the period is practically unchanged by light damping, okay?

4:58Graphs of the decay

Now look at the middle graph, capital A squared against time, a falling exponential curve. In equal intervals of time the energy falls by the same fraction, not the same amount. Got it? The graph on the right takes the natural log of capital A squared, and the curve becomes a straight line of negative slope, marked minus b over m. Clear? The energy is halved in a time t half, the small t with one over two below it. It equals m over b, times the natural log of two, and the dashed lines mark it. So how long does the amplitude take to halve? The amplitude takes twice this time, because halving the amplitude quarters the energy, and that takes two of these halving times. Got it?

5:44Example 1: find g from 20 oscillations in 40.2 s

Now let us do Example one. The effective length of a pendulum is one point zero zero metres, and it makes twenty oscillations in forty point two seconds. We have to find g. Pause the video and try it first. Are you done? Okay, let us check. One oscillation takes forty point two over twenty, which gives two point zero one seconds. Right? So g equals four pi squared times one point zero zero, over two point zero one squared. Work it out, and you get nine point seven seven metres per second squared.

6:21Example 2: percentage error in g from length and timing

Now the harder one, Example two. The length of a pendulum is fifty point zero centimetres, measured to one millimetre. Fifty oscillations take seventy one point zero seconds, on a watch reading to zero point one seconds. We have to find the percentage error in g. Pause the video and try it first. Are you done? Let us take the length first. One millimetre is zero point one centimetres, so delta L over L is zero point one over fifty point zero, or zero point two per cent. Clear? Now seventy one point zero over fifty gives capital T as one point four two seconds. So is delta capital T over capital T simply zero point one over one point four two? No, that is the trap, because the error is also divided by n, the number of oscillations. Delta capital T is delta t over n, and capital T is t over n, so the n cancels. The ratio is delta t over t, or zero point one over seventy one point zero, okay? Now put both into the error formula, and remember the two. Zero point one over fifty point zero, plus two times zero point one over seventy one point zero. That gives zero point two per cent, plus zero point two eight per cent, so about zero point five per cent. If you avoided that trap, very well done.

7:47Quick revision: g formula, L vs T squared, energy decay

So, quick revision. First, g equals four pi squared times L, over capital T squared, and L is thread, plus hook, plus radius of the bob. Second, L against capital T squared is a straight line through the origin, and the error in capital T counts double. Third, the energy goes with capital A squared, so it halves in half the time the amplitude takes. You now have this whole experiment in your hands. Now try the practice problems on this page, and all the best to you.