Energy in SHM and Small Oscillations

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle first. Picture a block on a spring, swinging between plus A and minus A, where A is the amplitude. Catch it halfway out, at x equals A over two. Is its energy half kinetic and half potential there? Halfway, so half and half, sounds fair, right? Hold on to that. Today we study energy in simple harmonic motion, and the period of small oscillations. This comes up often in JEE, so let's begin.

0:33Energy at every point of the motion

So, take a block of mass m on a spring with spring constant k. At a displacement x from the mean position, the spring stores potential energy capital U, equal to half k times x squared; the small two up top means squared, okay? It is zero at the mean position and largest at the two ends. Clear? Now the kinetic energy, capital K, okay? Let me ask you this: at the two ends, x equals plus or minus A, how fast is the block moving? Zero speed, exactly! The block stops for a moment at each end. Capital K is half m times v squared, with v the speed, and the speed formula from the last lesson makes it half k, times the bracket A squared minus x squared. So capital K is zero at the ends and largest at the mean position. Got it? Now add the two, and the x squared terms cancel! So capital E, the total energy, is half k times A squared, which is also half m, times omega squared, times A squared. Omega, the Greek letter like a curly w, is the angular frequency. No x is left, so the total energy is constant, the same at every point. With me?

1:44Where is kinetic equal to potential?

Now, back to our puzzle. Where are the kinetic and potential energies equal? Halfway out, at x equals A over two, right? Let's check. Set capital K equal to capital U, and the half k cancels on both sides. So we get A squared minus x squared equals x squared, okay? So two x squared equals A squared, and x equals plus or minus A over root two, not A over two! The tick-shaped sign is the square root. Now, A over root two is about zero point seven one times A, beyond halfway. So halfway out, the energy is not shared equally.

2:22The energy graph

Clear? Now let us see all of this on one graph. Energy goes up the side, and x along the bottom, from minus A to plus A, with the mean position O in the middle. The orange parabola is capital U, labelled half k times x squared: zero at O, and largest at the ends. Okay? The dark blue curve is capital K, labelled half k, times the bracket A squared minus x squared: largest at O, and zero at the ends. And the dashed red line on top is capital E, half k times A squared, flat, because the total never changes. Right? Now look at the dotted line, marked A over root two. That is exactly where the orange and blue curves cross, so capital K equals capital U there, just as we found. Got it?

3:07Kinetic and potential energy keep trading

Now watch the energy as time runs. During one full oscillation, how many times does capital U reach its maximum? Twice, yes, once at each end! So capital K and capital U each swing between zero and capital E with frequency two f, twice the frequency f of the block itself. Clear? Now, the angle brackets on screen mean average. Over one full cycle in time, the average of capital K equals the average of capital U, and each is capital E over two. Right? But average over position instead, and the answer changes. Why should that be? Here is the question: does the block spend more time near the middle, or near the ends? Near the ends, where it moves slowly and capital K is small. That pulls the time average down to half the total. Averaged over position, taking x evenly from minus A to plus A, the average of capital K is two thirds of capital E, okay? So in a problem, always check which average it asks for, over time or over position. With me?

4:15Small oscillations in any potential

Now a powerful idea. Suppose the potential energy, capital U of x, has a minimum at x naught, x with a small subscript zero. There, the first derivative U prime is zero, and the second derivative U double prime is greater than zero; each little dash is a prime. Okay? Now a quick question. For our spring, capital U is half k times x squared. What is its second derivative? Just k, the spring constant! So the second derivative at the minimum plays the role of k. Small oscillations about x naught are simple harmonic, SHM, with k equal to U double prime at x naught. Only small oscillations, though. Clear? So omega equals the square root of U double prime at x naught, over m, and capital T, the period, is two pi over omega, with pi the familiar Greek letter. Oh no, careful here: students often forget the mass m. With me? And adding a constant to capital U changes nothing, since its derivatives stay the same. Right?

5:16Example 1: total energy, K and U of a spring at x = A/2

Now, our puzzle with real numbers. Example one: a zero point five kilogram mass, on a spring of constant fifty newtons per metre, oscillates with amplitude zero point one metre. Find the total energy, and the kinetic and potential energies at x equals zero point zero five metre. On screen, k g is kilograms, m is metres, and capital N, m to the power minus one, is newtons per metre. Pause the video and try it first. Done? Okay, let's check. The total energy is half k times A squared: half times fifty times zero point one squared, which is zero point two five joules, where capital J is joules. Okay? Now at x equals zero point zero five metre, capital U is half k times x squared: half times fifty times zero point zero five squared, which gives zero point zero six two five joules. Right? And capital K is what is left: zero point two five minus zero point zero six two five, which is zero point one eight seven five joules. Clear? And notice, zero point zero five metre is half of zero point one metre: halfway out, our puzzle point! There, capital K is three times capital U, three quarters of the total energy. So three quarters kinetic and one quarter potential, not half and half, and our puzzle is solved. Got it?

6:38Example 2: equilibrium and omega for U = a/x2 - b/x

Now a harder one. Example two: a particle of mass m has potential energy capital U of x equals small a over x squared, minus b over x, with small a, b and x all positive. Find the equilibrium position, and the angular frequency of small oscillations. Pause the video and try it first. Done? Okay, let's check. At equilibrium the force is zero, so U prime is zero: minus two times a, over x cubed, plus b over x squared, equals zero. That gives x naught equals two times a, over b; the double arrow means implies, okay? But wait a moment: is this point a minimum of the potential, or a maximum? What must we check before we talk about oscillations? The second derivative, yes! U double prime of x is six times a, over x to the power four, minus two times b, over x cubed. Right? Now put in x naught, over the common denominator x naught to the power four. The top is six times a, minus two times b times x naught. Since b times x naught is two times a, the top is just two times a. With x naught equal to two times a, over b, this becomes b to the power four, over eight times a cubed, greater than zero, so it is a minimum. With me? Finally, omega is the square root of U double prime at x naught, over m: the square root of b to the power four, over eight times a cubed times m. See, the mass m sits in the denominator, just as we warned. Clear?

8:18Quick revision: K = U at A/root 2, 2f, omega of small swings

So, quick revision. The total energy, half k times A squared, is constant, and kinetic equals potential at plus or minus A over root two, not at A over two. Kinetic and potential energy oscillate at twice the frequency of the motion, and over time each averages half the total. For small oscillations about a minimum, omega is the square root of U double prime at x naught, over m. Now go and try the practice problems on this page. All the best!