9702 · 17.2Energy in simple harmonic motion

Energy exchange in SHM

0:00Pull a child's swing back and let it go. At the very top of each swing it stops for an instant. At the bottom it's moving fastest. Then it slows down, stops again on the other side, and comes back. If nothing slowed it down, it would keep reaching the same height forever. Something is being passed back and forth, but never lost. That something is energy. In this video we'll follow the energy of an oscillator through one cycle, derive the formula for its total energy, and see why doubling the amplitude multiplies the energy by four.

0:30From the last video: in simple harmonic motion, a equals minus omega squared x, and timing from the centre, x equals x-nought sin omega t, where x-nought is the amplitude and omega is the angular frequency. We also need the velocity. It's v equals v-nought cos omega t, where the maximum speed is v-nought equals omega x-nought. And the speed at any displacement is v equals plus or minus omega times the square root of x-nought squared minus x squared.

0:59We'll assume no friction or air resistance throughout. Energy loss is the subject of the next video, on damping.

Kinetic and potential energy transfer

1:06Let's start with the simplest oscillator: a mass on a horizontal spring, sliding on a frictionless surface. Pull it to one side by a distance x-nought and let go. At the extreme position, the block is stationary, so its kinetic energy is zero. But the spring is stretched as far as it ever gets, so it stores its maximum elastic potential energy. All the energy is potential. The spring pulls the block back towards the centre.

1:30The block speeds up as the stretch shrinks: potential energy is converted into kinetic energy. At the centre, the spring is at its natural length and stores no elastic potential energy, and the block moves at its maximum speed, v-nought, so all of the energy is kinetic energy. The block overshoots and squashes the spring, which pushes back and slows it down. Kinetic energy turns back into potential energy, until the block stops for an instant at the other extreme.

1:55Then it all runs in reverse. That's the interchange between kinetic and potential energy. Twice every cycle the energy is all potential, at the two extremes, and twice it's all kinetic, at the centre, once going each way. And the crucial point: with no friction, the total, kinetic plus potential, never changes. The energy just sloshes back and forth between the two forms. A simple pendulum swinging through small angles does exactly the same thing.

2:20The only difference is that the potential energy is gravitational. At each end, the bob is highest and momentarily stationary: zero kinetic energy, maximum potential energy. At the bottom, it's lowest and fastest: maximum kinetic energy, and, measuring potential energy from that lowest point, zero potential energy. So for any oscillator: kinetic energy is biggest at the centre, and potential energy is biggest at the extremes.

Deriving total SHM energy

2:47Now let's put a number on the total energy. The trick is to calculate it at the easiest moment: the centre, where potential energy is zero. There, the total energy is just the maximum kinetic energy, so E equals one-half m v-nought squared. Substitute v-nought equals omega x-nought, and v-nought squared becomes omega squared x-nought squared. So: E equals one-half m omega squared x-nought squared. This is the total energy of a system undergoing simple harmonic motion, and you need to recall and use it.

3:20Because the total is constant, it holds at every point in the cycle, not just at the centre. It depends on the mass, on omega squared, and on the amplitude squared. That last part is the big one: total energy is proportional to amplitude squared. Double the amplitude, and the energy goes up four times. Halve it, and the energy drops to a quarter. Why? For the same omega, twice the amplitude means twice the maximum speed, because v-nought equals omega x-nought.

3:46And kinetic energy depends on speed squared, so twice the speed means four times the energy.

Energy at displacement and graphs

3:53What about somewhere in between, at a displacement x? Start with kinetic energy, one-half m v squared. Square the speed equation, v equals plus or minus omega times the square root of x-nought squared minus x squared. Both the plus-or-minus and the square root disappear, leaving v squared equals omega squared times x-nought squared minus x squared. So the kinetic energy at displacement x is one-half m omega squared, times x-nought squared minus x squared.

4:23As a quick check, at the centre x is zero, and this becomes the total energy. At the extremes, x equals x-nought, and it becomes zero. Good. Potential energy is simply the total minus the kinetic energy. Subtract, and the x-nought squared terms cancel, leaving: Potential energy equals one-half m omega squared x squared. It's zero at the centre and grows with the square of the displacement, reaching the total energy at the extremes.

4:51These formulas give the first important graph: energy against displacement, from minus x-nought to plus x-nought. The total energy is constant, so it's a horizontal line. Potential energy, one-half m omega squared x squared, is a U-shaped parabola, zero at the centre and rising to meet the total energy line at both ends. Kinetic energy is the total minus the potential energy, so it's an upside-down parabola, like an arch: equal to the total at the centre, and zero at both ends.

5:19At any displacement, the heights of the two curves add up to the total. And the graph is symmetrical, because displacement is squared: plus x and minus x give the same energies. Where do the curves cross? There, kinetic energy equals potential energy, so each is half the total. Set one-half m omega squared x squared equal to half of one-half m omega squared x-nought squared. The constants cancel, giving x squared equals x-nought squared over two, so x equals x-nought over root two, about 0.71 of the amplitude.

5:53That surprises people. At half the amplitude, potential energy is only a quarter of the total, because displacement is squared. The energy is shared equally further out. The second graph is energy against time, timing from the centre, so x equals x-nought sin omega t. At time zero, kinetic energy is maximum and potential energy is zero. A quarter of a period later, the object is at an extreme: kinetic energy zero, potential energy maximum.

6:20After half a period, it's back at the centre moving the other way, and kinetic energy is maximum again. Here's the key feature. Kinetic energy peaks every time the object passes through the centre, which happens twice per oscillation. Potential energy peaks at both extremes, also twice per oscillation. So kinetic and potential energy each vary at twice the frequency of the oscillation, with half its period. You can see why from v equals v-nought cos omega t.

6:48When the velocity is negative, v squared is still positive, so every negative loop of the velocity graph becomes a positive hump of kinetic energy. Twice as many humps, and neither energy ever dips below zero. At every instant they add up to the constant total.

Worked example: oscillator energy

7:03Let's put all of this to work. A block of mass 0.20 kilograms is attached to a spring and oscillates on a smooth horizontal surface with simple harmonic motion. Its amplitude is 5.0 centimetres and its angular frequency is 20 radians per second. Find: one, the total energy of the oscillation. Two, the kinetic energy and potential energy when the displacement is 3.0 centimetres. And three, the displacement at which the kinetic energy equals the potential energy.

7:33First, convert units. The amplitude is 5.0 centimetres, which is 0.050 metres. The displacement of 3.0 centimetres is 0.030 metres. Part one: the total energy. E equals one-half m omega squared x-nought squared. That's one-half times 0.20, times 20 squared, which is 400, times 0.050 squared, which is 0.0025. One-half times 0.20 is 0.10, and 0.10 times 400 gives 40. Then 40 times twenty-five times ten to the minus four gives 0.10, so the total energy is 0.10 joules.

8:17Let's check that another way. The maximum speed is v-nought equals omega x-nought, which is 20 times 0.050, which is 1.0 metre per second. Maximum kinetic energy is one-half m v-nought squared, which is one-half times 0.20 times 1.0 squared, which is 0.10 joules. Same answer, as it must be. Part two: at a displacement of 0.030 metres. Potential energy is one-half m omega squared x squared. That's 40, which we already worked out for one-half m omega squared, times 0.030 squared, which is nine times ten to the minus four.

8:58So the potential energy is 0.036 joules. Kinetic energy is the total minus the potential energy: 0.10 minus 0.036, which is 0.064 joules. We can check the kinetic energy directly. The speed at x equals 0.030 metres is omega times the square root of x-nought squared minus x squared. In units of ten to the minus four, x-nought squared is twenty-five and x squared is nine, so the difference is sixteen times ten to the minus four.

9:30The square root of that is 0.040. So the speed is 20 times 0.040, which is 0.80 metres per second. Then kinetic energy is one-half times 0.20 times 0.80 squared, which is 0.10 times 0.64, which is 0.064 joules. It matches. Notice what that tells us. At 3 centimetres, which is 60 percent of the amplitude, the block has only 36 percent of the energy stored as potential energy, because 0.6 squared is 0.36.

10:05The remaining 64 percent is still kinetic. Part three: where is kinetic energy equal to potential energy? If the two are equal, each one must be half of the total energy, so the potential energy is 0.050 joules. Set 40 times x squared equal to 0.050. Then x squared is 0.050 divided by 40, which is twelve point five times ten to the minus four, and x is the square root of that, which is about 0.035 metres, or 3.5 centimetres.

10:37That's the same as x-nought over root two: 5.0 divided by root two is 3.5 centimetres. And since displacement can be either side of the centre, the answer is plus or minus 3.5 centimetres.

Common mistakes and exam practice

10:51Let's go through the mistakes that cost marks most often. Mistake one: thinking energy is proportional to amplitude. It's proportional to amplitude squared. Halve the amplitude and the energy becomes a quarter, not a half. Mistake two: forgetting to convert centimetres to metres. Because the amplitude is squared, that slip makes the energy wrong by a factor of ten thousand. Mistake three: saying kinetic energy has the same frequency as the oscillation.

11:18Kinetic and potential energy both vary at twice the frequency, because each peaks twice per cycle. Mistake four: mixing up which energy is maximum where. Picture the object. At the extremes, it's momentarily at rest, so kinetic energy is zero. At the centre, it's fastest, so kinetic energy is maximum. Questions on this topic usually test three things: using E equals one-half m omega squared x-nought squared, describing the energy changes in words, and interpreting energy graphs.

11:51Here's a typical example. "A simple pendulum oscillates with simple harmonic motion with a period of 2.0 seconds. One, describe the energy changes of the bob as it moves from one extreme position to the other. Two, state the period of the variation of the kinetic energy of the bob. And three, the amplitude of the oscillation is reduced to half its original value. State the fraction of the original total energy that the pendulum now has." Part one.

12:17A good answer names the energies, the positions, and the direction of conversion. At the extreme, the bob is momentarily at rest, so its kinetic energy is zero and its gravitational potential energy is maximum. As it swings down to the centre, potential energy is converted into kinetic energy, and at the centre kinetic energy is maximum. As it rises to the other extreme, kinetic energy is converted back into potential energy.

12:42And the total energy stays constant throughout. Part two. The kinetic energy is maximum every time the bob passes through the centre, which happens twice in each 2.0 second oscillation. So the period of the kinetic energy variation is 1.0 second. The trap here is writing 2.0 seconds. For part three, the total energy is proportional to the amplitude squared. Half the amplitude gives one-half squared, which is one-quarter. So the pendulum now has one-quarter of its original energy.

13:11The period doesn't come into it, because for SHM the period doesn't depend on the amplitude. Let's bring it all together. In undamped simple harmonic motion, energy is continuously exchanged between kinetic and potential energy, while the total stays constant. At the two extremes of the motion, the kinetic energy is zero and the potential energy is maximum. At the centre, the kinetic energy is maximum and the potential energy is zero.

13:36The total energy equals the maximum kinetic energy, which gives E equals one-half m omega squared x-nought squared. So total energy is proportional to amplitude squared. At displacement x, kinetic energy is one-half m omega squared times x-nought squared minus x squared, and potential energy is one-half m omega squared x squared. Against displacement, they form two parabolas under a flat total line, crossing at x-nought over root two. Against time, each oscillates at twice the frequency of the motion.

14:07Next time, we'll let friction back in and see what happens when an oscillator is damped.