WEBVTT
00:00:01.320 --> 00:00:04.400
In this video, we’re going to learn about damped oscillations.
00:00:04.840 --> 00:00:09.560
We’ll see the different types of damping that an oscillating system can experience.
00:00:10.000 --> 00:00:13.120
And we’ll also learn about energy in an oscillating system.
00:00:13.960 --> 00:00:22.080
To start out, imagine you have been entrusted with an egg from one of the rarest species of birds in the entire world, the California condor.
00:00:22.800 --> 00:00:29.840
Your mission, should you choose to accept it, is to transport this egg safely to a facility 100 kilometers away.
00:00:30.480 --> 00:00:36.520
The challenge in all this is that the way to the facility goes over rough ground where there is no paved road.
00:00:37.200 --> 00:00:43.320
In order to transport the egg without breaking it, the shocks on your vehicle will need to be precisely tuned.
00:00:43.880 --> 00:00:50.920
To understand just how the shocks should be tuned, it will be helpful to know something about damped oscillations.
00:00:50.920 --> 00:00:53.880
Damping is something that happens to oscillating systems.
00:00:54.400 --> 00:00:59.600
Damping resists the oscillations and tries to bring the system back to rest at equilibrium.
00:01:00.320 --> 00:01:04.120
Say, for example, that you hit a drum cymbal with a drumstick.
00:01:04.680 --> 00:01:10.760
The cymbal will resonate loudly in response to this contact and actually oscillate up and down.
00:01:11.600 --> 00:01:16.080
One way to stop those oscillations is to grab onto the cymbal with our hand.
00:01:16.440 --> 00:01:23.400
When we do this, we’re effectively damping that system; that is, we’re bringing it to rest at equilibrium.
00:01:23.400 --> 00:01:26.280
Oscillating systems can take many different forms.
00:01:26.720 --> 00:01:33.880
We could have a cymbal that’s vibrating or a pendulum that swings back and forth or the siren on an ambulance.
00:01:34.280 --> 00:01:36.160
All of these are systems that oscillate.
00:01:36.800 --> 00:01:43.040
The system we’ll focus on for the next few minutes though is the oscillating system of a mass on the end of a spring.
00:01:43.640 --> 00:01:55.200
If we were to stretch the spring, which has a spring constant of 𝑘, past its equilibrium point and then release the spring, by Hooke’s Law there would be an upward acting force on the mass.
00:01:55.600 --> 00:02:00.800
Neglecting the effects of gravity, this would then be the only force that the spring is subject to.
00:02:01.400 --> 00:02:13.480
If we were to write out Newton’s second law for the forces acting on this oscillating system, we would simply have the spring force, negative 𝑘𝑥, being equal to the mass of the system times its acceleration.
00:02:14.280 --> 00:02:17.720
But now, imagine we were to introduce another force to our spring.
00:02:18.320 --> 00:02:24.520
Imagine that, as our mass was moving down, we introduce a damping force which counteracts that motion.
00:02:25.160 --> 00:02:30.560
This damping force would be proportional to and opposite the velocity of our moving mass.
00:02:31.040 --> 00:02:35.480
The idea with the damping force is that it resists the motion of our oscillator.
00:02:36.120 --> 00:02:41.960
Including this new force in our second law expression, we now have a damped oscillating system.
00:02:42.680 --> 00:02:49.280
We can call this factor of negative 𝑐 times 𝑣, where 𝑐 is a constant, the damping force.
00:02:49.280 --> 00:02:57.040
And what’s more, we can specify a damping coefficient, which is equal to our constant 𝑐 divided by twice the mass of our oscillator.
00:02:57.760 --> 00:03:01.520
This coefficient is often represented using the Greek letter 𝛾.
00:03:02.480 --> 00:03:05.880
Let’s consider the effects of damping on an oscillating system.
00:03:06.560 --> 00:03:11.600
Imagine that we plot the position of the mass on the end of our oscillator as a function of time.
00:03:12.440 --> 00:03:23.600
We’ve seen that, for a simple harmonic oscillating system, position as a function of time is equal to the amplitude of that system multiplied by the sin of its angular frequency times time.
00:03:24.320 --> 00:03:29.760
For an undamped oscillator, our systems oscillations might look like this over time.
00:03:30.320 --> 00:03:35.840
And if no frictional force resisted the oscillations, the system would look like this indefinitely.
00:03:36.120 --> 00:03:40.120
It would keep oscillating up and down or back and forth or however it is moving.
00:03:40.880 --> 00:03:47.240
But what if we add a damping force and with a damping force we add a damping coefficient?
00:03:47.840 --> 00:03:49.520
What will this do to our curve?
00:03:50.120 --> 00:03:55.080
Well, we can imagine that if we damp a system, it will slowly approach zero.
00:03:55.080 --> 00:04:04.480
A mathematical function that we can tune to give us a slow asymptotic approach to a particular limiting value is the exponent function 𝑒.
00:04:04.480 --> 00:04:15.840
In our case, we would say 𝑒 to the negative 𝛾 multiplied by time, because the more time elapses, the more our system will tend towards zero, its equilibrium location.
00:04:16.560 --> 00:04:25.880
To include the effects of damping in our system’s motion is to include this exponential term in our equation for position as a function of time.
00:04:26.560 --> 00:04:32.600
The way that our position depends on this exponential factor has to do with the value we have for 𝛾.
00:04:33.320 --> 00:04:41.000
Let’s see that we shift our curve by a phase difference so that now it begins when 𝑡 is equal to zero at its maximum amplitude value.
00:04:41.640 --> 00:04:46.600
As we tune our value for 𝛾, the damping coefficient, let’s say that we have a goal.
00:04:47.160 --> 00:04:55.560
Our goal is for this mass on an oscillating system to return as quickly as possible to its equilibrium position 𝑥 and stay there.
00:04:56.280 --> 00:05:03.400
In support of that goal, we choose a particular value for 𝛾 inserted into our equation and then plot that curve on our graph.
00:05:03.960 --> 00:05:06.280
And say that the curve looks like this.
00:05:06.880 --> 00:05:11.360
So we see that our damping is having an impact on the motion of our system.
00:05:11.960 --> 00:05:19.520
The only thing is the system keeps moving back through its zero point where we would like it to come to a rest, but it doesn’t stop there.
00:05:19.520 --> 00:05:20.320
It keeps going.
00:05:21.240 --> 00:05:29.440
We see that eventually the system will reach and stay at its equilibrium point if this trend continues, but we like it to happen more efficiently.
00:05:30.240 --> 00:05:38.080
When our position versus time curve looks like this, we say that the system is underdamped or that we are underdamping the system.
00:05:38.720 --> 00:05:44.400
This means that, by increasing 𝛾, we can get the system to reach equilibrium more quickly and efficiently.
00:05:45.080 --> 00:05:50.360
Say that we do just that; we go back and we turn up our value for 𝛾 to a higher value.
00:05:50.920 --> 00:05:57.840
Say that when we do and we plot that on our position versus time curve, we get a curve as shown in the green dashes.
00:05:58.400 --> 00:06:01.440
This curve certainly looks different from our underdamped result.
00:06:02.080 --> 00:06:11.320
Rather than crossing over our zero point for equilibrium many times as in our underdamped solution, this one just reaches zero once and then stays there.
00:06:11.960 --> 00:06:15.000
That’s good, but notice one other thing about this curve.
00:06:15.480 --> 00:06:27.920
If we compare how fast this curve reaches equilibrium compared to the first equilibrium point of our underdamped system, we see there’s a significant gap in time between those two points.
00:06:27.920 --> 00:06:34.880
In our ideal case, not only would we reach zero and then not overshoot it, but we would reach zero as quickly as possible.
00:06:35.720 --> 00:06:40.480
This green curve, it turns out, involves a value of 𝛾 which is higher than what we would want.
00:06:41.000 --> 00:06:43.200
It’s an example of what’s called overdamping.
00:06:43.840 --> 00:06:51.560
As a picture of overdamping, imagine a pendulum that naturally swings back and forth which we then apply a damping force to.
00:06:52.200 --> 00:07:06.040
If the damping force was excessive — that is, if the constant 𝑐 that we chose was too high — then our pendulum will take an unnaturally long amount of time to ever so slowly return to its equilibrium position.
00:07:06.760 --> 00:07:13.760
It does get there and stay there once it does, but it takes so long to arrive that we call this overdamping our system.
00:07:14.400 --> 00:07:19.360
As you might guess, there’s a happy medium that exists between under and overdamping our system.
00:07:19.960 --> 00:07:23.720
And to find it, we go back to our choice of 𝛾.
00:07:23.720 --> 00:07:35.080
If we pick a value of 𝛾 between the two values we’ve chosen previously and then insert that into a position versus time function, we get a curve that looks like this blue dashed line.
00:07:35.640 --> 00:07:45.000
In this case, our system does reach zero and stay there and it does it fairly quickly, not quite as quickly as the underdamped system, but fairly quickly nonetheless.
00:07:45.400 --> 00:07:51.800
In this circumstance, our system is said to be critically damped or that we have a critical damping value for 𝛾.
00:07:52.520 --> 00:07:53.880
This is the sweet spot.
00:07:54.080 --> 00:07:56.640
We’re neither over- nor underdamping our system.
00:07:57.480 --> 00:08:06.360
When 𝛾 is set to be equal to the resonant frequency of our oscillating system, that’s when we achieve this critical damping curve.
00:08:06.920 --> 00:08:15.480
For a mass on an oscillating spring, that means that 𝛾 is equal to 𝜔, which is the square root of the spring’s constant over its mass.
00:08:16.160 --> 00:08:23.560
By setting 𝛾 equal to that value depending on the physical parameters of our system, we achieve the critical damping we want.
00:08:24.280 --> 00:08:29.280
Keep in mind that this particular value for critical damping has to do with our particular system.
00:08:29.920 --> 00:08:33.400
If our system were an oscillating pendulum, that value would be different.
00:08:34.040 --> 00:08:37.920
It would instead in that case be equal to the square root of 𝑔 over 𝑙.
00:08:38.560 --> 00:08:45.080
Now, we know that a damped oscillator can be underdamped, it can be over damped, or it can be critically damped.
00:08:45.560 --> 00:08:48.400
And that all depends on the value that we choose for 𝛾.
00:08:48.400 --> 00:08:53.280
Now, let’s turn our attention to energy in an oscillating system.
00:08:53.960 --> 00:09:01.360
Consider another scenario with a mass on an oscillating spring, where our mass is displaced from its equilibrium location.
00:09:01.560 --> 00:09:03.600
And it’s moving downward with a speed 𝑣.
00:09:04.080 --> 00:09:06.040
And the spring has a spring constant 𝑘.
00:09:06.720 --> 00:09:12.720
If we consider the total energy of the system, we know it’s the sum of the system’s kinetic plus potential energy.
00:09:13.400 --> 00:09:29.200
If the oscillation amplitudes of this mass are small, then we can neglect gravitational potential energy and our potential energy of the system entirely consists of spring potential energy, one-half 𝑘 times 𝑥 squared, where 𝑥 is our displacement from equilibrium.
00:09:29.800 --> 00:09:34.040
The kinetic energy of the system is equal to one-half mass times the speed squared.
00:09:34.520 --> 00:09:47.200
Since our mass 𝑚 moves in simple harmonic motion, its position 𝑥 as a function of time is given as we’ve seen by the product of its amplitude times the sin of its angular frequency times time.
00:09:47.880 --> 00:09:53.440
And in addition, the speed of our mass 𝑣 is the time derivative of position as a function of time.
00:09:54.120 --> 00:10:00.640
The time derivative of 𝐴 sin 𝜔𝑡 is 𝜔 times 𝐴 times the cos of 𝜔𝑡.
00:10:01.200 --> 00:10:08.400
So we take our value for position and our value for speed and insert them into our equation for overall system energy.
00:10:09.240 --> 00:10:11.960
We now have an expression for the total energy of our system.
00:10:12.400 --> 00:10:17.120
And we can simplify it further by considering the factor of 𝑚 times 𝜔 squared.
00:10:17.840 --> 00:10:32.080
Recalling that 𝜔, the angular frequency of a mass on a spring, is equal to the square root of the spring constant over the mass, when we square both sides, we see that 𝜔 squared is equal to 𝑘 over 𝑚.
00:10:32.080 --> 00:10:44.080
Substituting that into our equation for kinetic energy, we see that the factor of mass cancels out, giving us an expression where both terms involved have factors of one-half 𝑘 times 𝐴, the amplitude, squared.
00:10:44.680 --> 00:10:51.600
Factoring that term out, we see it multiplies the quantity cos squared 𝜔𝑡 plus sin squared 𝜔𝑡.
00:10:52.120 --> 00:11:05.520
There is a trigonometric identity which says that the sin squared of any quality plus the cos squared of that same quantity is equal to one, meaning that the entire expression inside our parentheses is equal to one.
00:11:06.440 --> 00:11:11.120
We’ve arrived then at an expression for the total energy in an oscillating spring system.
00:11:11.640 --> 00:11:15.440
It’s one-half the spring constant 𝑘 times the amplitude of the system squared.
00:11:16.120 --> 00:11:20.320
Even in a damped system, this energy amount can be useful to us as an initial energy value of our oscillator.
00:11:20.320 --> 00:11:22.360
Let’s review what we’ve learned so far about damped oscillations.
00:11:23.200 --> 00:11:27.000
Let’s review what we’ve learned so far about damped oscillations.
00:11:28.280 --> 00:11:33.920
We’ve seen that damping is when an oscillating system is brought to rest at its equilibrium position.
00:11:34.800 --> 00:11:42.320
We’ve also seen that a system can be under, over, or critically damped depending on the damping coefficient 𝛾.
00:11:43.000 --> 00:11:50.280
Critical damping happens when our damping coefficient 𝛾 is equal to the resonant frequency 𝜔 for system.
00:11:50.960 --> 00:11:58.680
For a mass on an oscillating spring, this value for 𝜔 is equal to the square root of the spring constant divided by the mass.
00:11:59.400 --> 00:12:10.520
And finally, we saw that the total energy in a harmonically oscillating spring system is equal to one-half the spring constant 𝑘 times the amplitude of oscillation squared.