The Simple Pendulum

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. You hang a small pendulum from the roof of a car, and the car starts speeding up steadily. I have two questions for you. Where does the bob hang now, and does it swing faster or slower than before? Pendulums in lifts and cars come up often in JEE papers. So, let's begin.

0:22Why a small swing is simple harmonic

Picture a bob of mass m on a string, pulled aside by an angle theta, the Greek letter, from the vertical. F, the restoring force along the arc, equals minus m times g times sine theta; the minus sign means it points back. For a small angle in radians, sine theta is about theta; that wavy equals sign means approximately. With me? Let me ask you this. What does a motion need, to be simple harmonic? Yes, exactly, a restoring force proportional to the displacement! Along the arc, the bob's displacement, called x, equals l times theta, where l is the length of the string. So the force is about minus m g over l, times x, proportional to x and pointing back. So small swings are SHM, with a spring constant k equal to m g over l. Okay? Now here is a question for you. Suppose the bob is twice as heavy. It is harder to move, so it swings more slowly, right? That is the trap, because it does not slow down at all, and here is why. From the spring mass lesson, T equals two pi times the root of m over k. Put k equals m g over l, and the mass cancels. So T equals two pi times the root of l over g. The heavier bob is harder to move, but it also feels a bigger pull. Clear? So the period depends only on l and g, not on the mass. It is also nearly independent of the amplitude, but only for small swings, where sine theta stays close to theta. Got it?

2:02Lifts and cars: use the effective g

Now, put the pendulum in an accelerating lift, car or trolley. Replace g with g effective, written with the small subscript e f f, so T equals two pi times the root of, l over g effective. The little arrows mean vectors, and a sub zero is the frame's acceleration. So g effective is the vector g, minus the vector a sub zero. Okay? And if an extra constant force, capital F, acts on the bob, add capital F over m to g instead. Same idea, right? Let me ask you this. The lift accelerates upwards. Does the pendulum swing faster, or slower? It swings faster, exactly! With the lift accelerating up at small a, minus a sub zero points down, so g effective is g plus small a. Bigger g effective, smaller period. With me? Accelerating downwards, it is the opposite. g effective is g minus small a, so the period becomes longer. Okay? Now suppose the cable snaps, and the lift falls freely. What happens to our pendulum? In free fall, small a equals g, so g effective is zero, and T goes to infinity; that sideways eight is the infinity sign. There is no oscillation at all, since nothing pulls the bob back. Clear? And in a car with horizontal acceleration small a, gravity points down while minus a sub zero points backwards, at right angles. So g effective is the root of, g squared plus small a squared, and the bob hangs along g effective, tilted backwards. Got it?

3:41Very long pendulums and pendulum clocks

Now, what if the string is not small compared with the Earth's radius? Then T equals two pi times the root of, one over, g times, one over l plus one over capital R, where capital R is that radius. Okay? So here is a question. If we keep making the string longer and longer, can the period become as large as we like? It cannot, and the formula shows why. As l goes to infinity, one over l goes to zero, and T goes to two pi times the root of capital R over g. That is about eighty four point six minutes, the largest possible. Right? Next, pendulum clocks. The Greek capital delta means a small change. Delta T over T equals half of delta l over l, minus half of delta g over g. With me? Suppose the period becomes a little longer. Does the clock gain time, or lose time? The clock loses time, because each tick takes longer. With eighty six thousand four hundred seconds in a day, the time lost or gained per day is that number, times delta T over T, clear?

4:52Example 1: period of a 1 m pendulum, seconds pendulum length

Example one, an easy one. Find the period of a pendulum one metre long, taking g as nine point eight metres per second squared, and then find the length of a seconds pendulum. Pause the video, and try it first. Done? Okay, let's check. T equals two pi times the root of l over g, so two pi times the root of, one over nine point eight. That comes to about two point zero one seconds. Okay? Now the seconds pendulum. Its name says seconds, so its period is one second, right? Oh no, careful, that is the trap! The seconds pendulum ticks once every second, so one swing across takes one second, and the full period, there and back, is two seconds. Got it? Square the period formula and solve: l equals g times T squared, over four pi squared. With T equals two, T squared is four, and the fours cancel, leaving g over pi squared. Nine point eight over pi squared gives about zero point nine nine three metres. That is why our one metre pendulum took a little over two seconds. Clear?

6:05Example 2: pendulum in a car accelerating at g over root 3

Now, remember our car puzzle? So, a pendulum hangs in a car accelerating horizontally at g over root three. Find the angle of its string with the vertical, and its new period in terms of its period T at rest. Pause the video, and try it first. Done? Okay, let's check. The string lies along g effective, with g pointing down and minus a sub zero pointing backwards. So tan theta equals small a over g, where small a is g over root three, giving one over root three; the double arrow means which gives. So theta is thirty degrees; the little circle means degrees. Okay? Now the period. The car moves sideways and gravity has not changed, so the period stays the same, right? No, no, careful, because the pendulum feels g effective, not g. g effective is the root of, g squared plus g squared over three, which is g times the root of four over three, so two g over root three. That is bigger than g itself, clear? T prime, the new period, equals T times the root of, g over g effective. That ratio is root three over two, so T prime equals T times, root three over two, to the power half. With me? Root three over two, to the power half, is about zero point nine three. So T prime is about zero point nine three times T, a shorter period. And our car puzzle is solved! For this car, the bob leans back by thirty degrees, and swings faster than before. Got it?

7:43Quick revision: period, effective g, seconds pendulum

So, quick revision. For small swings, T equals two pi times the root of l over g, whatever the mass. In a lift, a car or a trolley, use g effective, the vector g minus the frame's acceleration. And a seconds pendulum has a period of two seconds and a length of about zero point nine nine metres. Now try the practice problems on this page. All the best!