Damped and Forced Oscillations

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle first. Pull a swing back once and let it go. Every swing is a little lower, and slowly it stops, right? Now give it tiny pushes instead, each at just the right moment, and it goes higher and higher. Tiny pushes, and yet a huge swing? How is that possible? Today we study damped oscillations, which die away, and forced oscillations, kept going by a push. Both are part of JEE physics, so let's begin.

0:33Damping: the swings slowly die away

So, a swing dies down because something, like air, resists the motion. Let us take a resistive force, F equals minus b times v, with v the velocity and b the damping constant. The minus sign means it always opposes the motion. Okay? So, x equals A naught, times e to the power minus b times t over two m, times sine of omega prime times t, plus phi. Here x is the displacement, m is the mass, and A naught, the A with a small zero, is the starting amplitude. The letter e is the base of natural logarithms, omega prime, the omega with a little dash, is the angular frequency with damping, and phi, the Greek letter at the end, is the phase. Right? In the graph, x goes up the side and time t along the bottom. The dashed red curves are the envelope, A naught times e to the power minus b times t over two m. The exponential squeezes each swing smaller. Clear? Now, compare omega prime with omega naught, the omega with a small zero, which is the angular frequency without damping. Let me ask you this: with damping, does the system oscillate faster or slower? It oscillates slightly slower, yes, exactly! Omega prime is the square root of omega naught squared, minus the bracket b over two m, whole squared. We subtract something, so omega prime is smaller. Got it? And omega naught is the square root of k over m, with k the spring constant. So omega prime is slightly less than omega naught. With me?

2:11Amplitude and energy: how fast do they fall?

Alright, the amplitude goes as e to the power minus b times t over two m; the curly sign means proportional to. Set the exponential equal to one half, and take the natural log, which you see written as l n on screen. The amplitude halves at t equals two m, times natural log of two, over b. Okay? Now a quick question. The amplitude of a damped oscillator has halved. What fraction of its energy remains? If you said half, careful, that is the classic trap! Capital E, the energy, is proportional to the amplitude squared, so half the amplitude leaves only one quarter of the energy. Squaring the envelope doubles the exponent, so energy goes as e to the power minus b times t over m. Right? So the energy halves in m times natural log of two, over b, half the time the amplitude takes, so twice as fast. Clear? One more question. In each period, does the amplitude drop by the same amount, or by the same fraction? It drops by the same fraction, exactly! Each period multiplies it by the same number r, which is less than one. So after n periods, the amplitude is A naught times r to the power n. With me?

3:27Forced oscillations and resonance

Now back to our swing with the tiny pushes: that is a forced oscillation. The periodic driving force is F naught times sine of omega times t. Here F naught, the F with a small zero, is the largest value of the push, and this omega is the driving angular frequency. Right? Let me ask you this. Take a system with natural frequency fifty hertz, driven at forty hertz. At what frequency does it oscillate in the steady state? It oscillates at forty hertz, the driving frequency! If you said fifty, careful: in the steady state, a forced oscillation follows the driving force, at omega, not at omega naught. Okay? Now the amplitude. Capital A equals F naught, over the square root of m squared times the bracket omega naught squared minus omega squared, whole squared, plus b squared times omega squared. Clear? Here is the key question. For which driving omega does the bottom become smallest, making the amplitude biggest? When omega is close to omega naught, yes! Then the first bracket nearly vanishes, and the amplitude is largest. This is resonance, and the wavy equals sign means approximately. Got it? Near resonance, only the b squared times omega squared term really remains at the bottom, so a smaller damping b gives a taller peak. And the smaller the damping, the sharper the peak becomes too. With me? So our puzzle is solved. Tiny pushes, in step with the swing's natural frequency, give resonance. Tuning a radio uses resonance too. And soldiers break step on a bridge, because if their marching frequency matched a natural frequency of the bridge, resonance could build up a dangerously large amplitude. Clear?

5:18Example 1: damping constant b when amplitude halves in 2 min

Now, an example. Example one: a one kilogram block oscillates with amplitude twelve centimetres, and after two minutes the amplitude is six centimetres. Find the damping constant b, taking natural log of two as zero point six nine three. On screen, k g is kilograms and c m is centimetres. Pause the video and try it first. Done? Okay, let's check. Two minutes, m i n on screen, is one hundred and twenty seconds, with s for seconds, so careful, do not put t equal to two! And the arrow shows twelve centimetres going down to six, so the amplitude has halved. Okay? So e to the power minus b times t over two m equals one half. Taking the natural log, b times t over two m equals natural log of two; the double arrow means implies. Right? Put in m equals one kilogram and t equals one hundred and twenty seconds. One hundred and twenty over two gives sixty, so sixty times b equals zero point six nine three. Clear? So b is zero point six nine three over sixty, about one point one six times ten to the power minus two. The unit is kilograms per second; the minus one up top on the s means per second. That fits, because b times v must be a force. Got it?

6:43Example 2: amplitude and energy left after 5 damped swings

Now a slightly harder one. Example two: a pendulum loses ten percent of its amplitude in each oscillation. What fraction of the original amplitude remains after five oscillations, and what fraction of the energy? Pause the video and try it first. Done? Okay, let's check. Ten percent lost, five times over, makes fifty percent lost. So half the amplitude is left, right? No, no, careful, that is the trap! Each oscillation loses ten percent of what is there at that moment, not of the original. So each one keeps ninety percent, and r equals zero point nine. Okay? After five oscillations, the amplitude is A naught times zero point nine to the power five. That is about zero point five nine of the original. So about fifty nine percent is left, not fifty. Right? Energy goes as the amplitude squared, so squaring gives zero point nine to the power ten, which is about zero point three five. So fifty nine percent of the amplitude is left, but only about thirty five percent of the energy. Clear?

7:53Quick revision: damping decay, energy loss and resonance

So, quick revision. With damping, the amplitude goes as e to the power minus b times t over two m, the energy as e to the power minus b times t over m, so energy halves twice as fast. The amplitude loses the same fraction each period, so after n periods it is A naught times r to the power n. And a forced oscillator settles at the driving frequency, with resonance when omega is close to omega naught. Now go and try the practice problems on this page. All the best!