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Bo: Hey guys, Billy, Hey Bo, Bobby Hi Bo lyrics, Flipping Physics, Mr P, Ladies and gentlepeople. The bell has rung, therefore class has begun.
Therefore, you should be seated in your seat, ready and excited to do a basic example problem about acceleration and to understand a little bit about the direction for acceleration Bo Ready Bobby Set Billy Go Mr P.
Okay, here we go Billy.
Please read the problem.
Billy, Mr P is riding his bike at 14 3 kilometers per hour when he begins pedalling the bike to cause a constant acceleration.
If, after 6 4 seconds, the bike is moving at 23 7 kilometers per hour.
What was the acceleration of the bike Bobby Uh? Can you show that again, please That happened pretty fast, Mr P, For you Bobby yes Great Now let’s translate the problem to physics Billy.
If you could, please read again and Bo, could you please do the translation Billy? Mr P is riding his bike at 14 3 kilometers per hour when he begins pedaling Bo.
Please stop.
The velocity initial is 14 3 kilometers per hour. Billy Pedaling the bike to cause a constant acceleration If, after 6 4 seconds, the bike is moving at 23 7 kilometers per hour.
What was the acceleration of the bike Bo? The change in time is 6 4 seconds.
The final velocity is 23 7 kilometers per hour and acceleration equals question mark, because acceleration is what we’re trying to find Mr P Bobby what next Bobby.
Well, we learned last time that kilometers per hour and seconds don’t match very well.
So let’s convert both of the velocities to meters per second.
We know there are 3 600 seconds in an hour, so multiply both velocities by 1 hour over 3 600 seconds to cancel out the hours.
We also know that there are 1 000 meters in 1 kilometer, so multipy, both velocities by 1000 meters over 1 meter to cancel out the kilometers.
Mr P, What do we get then for numbers for velocity initial and velocity final Bobby? I actually have both answers.
The initial velocity is 3 972 and the 2 is repeating, and the final velocity is 6 583, with a repeating 3, Mr P, Great Bo.
What should we do now? Bo Again? We start with the equation: Billy Bobby The equation Bo. Yes, the equation Acceleration equals the change in velocity over the change in time where the change in velocity is equal to velocity final minus velocity initial.
Once we have the whole equation written down, we can just plug in all the numbers, because we have the numbers and we get 6 5833333 minus a 3 9722222 That whole quantity divided by 6.
4.
Mr P, Yes, it is that simple, and what do we get for numbers Bobby That worked out to be 0 4079861 with 2 sig figs? That would be 0 41 meters per second squared, Mr P, Yup.
We get 0 41 meters per second squared 2 sig figs because of the 6 4 seconds Questions Bo, I have 1 Bobby.
Actually I have 1 too, but you can go first Bo Thanks Bobby In the video the velocities weren’t actually negative.
Why not? Mr P, Oh yeah Good eye Bo.
You’re talking about the number shown on the bike right Bo Yup, Mr P.
Well, okay, Billy! What is the name of the item on the bike that is showing the number 23 7 kilometers per hour Billy? I think it’s called a speedometer.
Mr P, Correct. We’re reading that 23 7 kilometers per hour and the 14 3 kilometers per hour off of the speedometer on the bike Notice.
It’s a speedometer, not a velocitometer, A speedometer, not a velocitometer or whatever it would be called displays, just the speed, not the velocity.
So in this particular case it’s giving us the magnitude of the velocity at that particular time.
So the magnitude of velocity at this particular time is going to be the speed and not the velocity.
Now we do know it’s negative, because I’m moving to the left in the video Bo, Okay yeah.
That makes sense Thanks Bobby.
What was your question Bobby My question had, to do with the fact that the bike is.
Speeding up and if the bike is speeding, up shouldn,’t the, acceleration be, positive I, thought that negative accelerations meant that you were slowing down? Mr P, Oh yeah the direction.
I guess we should talk about that.
It is in the title of the video First off realize that we did do the problem correctly. We should end up with a negative acceleration, even though we’re speeding up.
Let,’s talk through why that.
Is by, way of some examples Billy Okay Bobby Good Let’s do that Mr P, Okay, let’s.
Say by way of an example, that I, 39, m riding, my.
Bike, to the right and speeding.
Up In this particular case, that would mean that my velocity would be to the right, and my acceleration would also be to the right and both would be positive because they would be to the right If instead, I’m riding my bike to the right and Slowing down my velocity again is going to be to the right and positive.
However, my acceleration is going to be to the left and negative Billy.
Oh, in other words, when you are speeding up, the acceleration is in the same direction as the velocity Bobby And when you’re slowing down, then the velocity and acceleration are in opposite directions.
Mr P, Exactly If the object is speeding up, that means that the velocity and acceleration are both going to be in the same direction.
However, if the object is slowing down, that means that the velocity and acceleration are going to be in opposite directions, which means, if I am riding my bike to the left and speeding up then, because I’m riding my bike to the left. The velocity is going to be to the left and negative.
If I’m speeding up, then my acceleration is going to be in the same direction as the velocity, which is going to be to the left and negative.
If, however, I am riding my bike to the left and slowing down, that means that my velocity is still going to be to the left and negative.
However, because I’m slowing down, the acceleration is going to be opposite of the direction of the velocity And therefore the acceleration is going to be to the right and positive Bobby Thanks, Mr P.
Now I do think that that bears repeating and writing down on the board Again, if the object is speeding up, the velocity and acceleration are in the same direction and if the object is slowing down the velocity and acceleration are in opposite directions.
Let’s take 1 more look at the 4 examples because darn it.
It took me a long time to film.
Well any more questions Bobby Nope Billy, No Bobby Thanks! Mr P.
Actually, I’m a bit surprised that there aren’t any more questions, because if you look at our acceleration, this is our answer for acceleration.
However, it doesn’t have a direction and acceleration has both magnitude and direction. Unfortunately, because of the problem statement, we didn’t know enough to be able to figure out a direction from the problem.
I kind of did this on purpose because you’re going to run into problems where you’re not given a direction and, alas, you can’t actually figure out a direction in your answer, and this is what you’re going to need to Do So I thought it would be good to go through an example like that, although in the problem they probably should have said something about direction so that you could figure out the direction for your answer, because again, acceleration has both magnitude and direction There you go.
Ladies and gentlepeople, I enjoyed learning with you today.
I hope you enjoyed learning with me.
Voiceover Lecture notes are available at FlippingPhysics com.
Please enjoy lecture notes responsibly, Voiceover You can both film.
At the same time, Child Yeah, Mr P, Hold this Child.
I don’t know how to work this one.
Mr P, It’s the same one, So move forward and stand about here, and I want you to film what mommy and I are going to do Child.
So do you hold it like this? Mr P? Yup, well, you can see what you’re filming Child. Well, I can’t get the whole car.
When do I press record, It is recording right now I’m filming you Child.
No, you’re not supposed to be Child.
I’m, not filming.
I don’t know how Child I got some video.
When do I press record, Mr P, So what I’ll do is I’ll yell 3.
2.
1 brake.
I’ll go over to the other side, So we’re going to do the same thing.
Only I’m going to be on this side of you now Why don’t we head back, I’m going to take some wearing the chest. Harness .
Thanks to Curiosity Stream for sponsoring
this video. A good understanding of how structures behave
when vibrating is what allows engineers to build rotating machinery, to launch sensitive
instruments into space, and to safely design buildings in seismic areas, to name just a
few of the many applications. Systems like these can be very complex, so
to study their vibrating behaviour engineers usually start by building a simple model that
approximates the dynamics of the system, but is easier to assess. The two most important parameters in any vibrating
system are its mass and its stiffness. A common modelling approach is to lump all
of the different contributions to mass and stiffness together, and represent them using
a point mass with a mass m and a spring with a stiffness k. This is called the lumped parameter modelling
approach. This kind of simplified model might seem quite
abstract, but can actually represent the dynamic behaviour of a lot of real systems quite accurately. The beauty of this simplicity is that we now
have something that we can analyse mathematically.
But first we have to make a few assumptions. We'll assume that the mass can only move up
and down. Since the system behaviour is defined by a
single output, the x coordinate of the mass, this is what's called a single degree-of-freedom
model. We'll also neglect the effects of gravity,
and for now we'll assume that there's no damping, meaning that no energy is lost from the system
as it vibrates, by friction or other means.
No external loads are acting on the system
– the purpose of the model is to understand how the system behaves in free vibration,
or in other words how it will oscillate when it's displaced, and then released. Since we've assumed there's no damping, the
mass will continue to oscillate like this indefinitely. The way the system vibrates is defined by
its equation of motion, which can be determined by applying Newton's second law. The second law states that the sum of the
forces acting on the point mass is equal to the product of its mass and its acceleration,
F=ma. We can figure out the sum of the forces acting
on the mass by drawing a free body diagram.
There is only one force, the force exerted
by the spring, which is equal to the displacement x multiplied by the spring stiffness k. And
so we obtain the equation of motion for the system. This is an ordinary differential equation,
and the solution is a sinusoidal function. t is time, Phi is the phase angle and A is
the amplitude of vibration. We can determine A and Phi by considering
the initial position and velocity of the mass. Let's look at an example where the system
has a mass of 5 kilograms and a spring stiffness of 20 Newtons-per-metre, and vibration is
triggered by applying an upwards velocity of 2 centimetres per second to the mass. Since the displacement x is initially zero,
the phase angle Phi must also be equal to zero. And then we can differentiate the equation
for x to calculate the amplitude of vibration A. An important property that can be calculated
from a mass-spring model is the system's natural frequency, the frequency at which it will
oscillate naturally when in free vibration. It's given by this term in the solution to the equation of
motion, and is denoted using the Greek letter Omega.
It depends only on the mass and the spring
stiffness, so no matter what the initial conditions are a system will always oscillate at the
same frequency. It has units of radians per second, so is
called the angular natural frequency. But it's sometimes more practical to think
of the natural frequency as a number of cycles per second, in which case it's denoted using
the letter f and has units of Hertz. The inverse of the natural frequency is the
period T, which is the duration of each cycle in seconds. Let's compare how two different systems oscillate. Both of these models have the same spring
stiffness, but different masses, and so different natural frequencies. The heavier mass oscillates at a much lower
frequency. A neat demonstration of the natural frequency
is the tuning fork. When the fork is struck it vibrates at its
natural frequency, which is much faster than shown here.
This causes the air molecules to vibrate at
that same frequency, which produces a corresponding tone. By assuming that the prongs behave like cantilever
beams in bending, beam theory can be used to derive a formula for the natural frequency
of the fork. The density, length and cross-section of the
prongs can be calibrated to obtain the desired tone. Of course when a mass oscillates freely it
doesn't do so indefinitely. Energy within the system is dissipated as
heat over time, so the oscillations progressively decrease in magnitude and eventually stop
altogether. This loss of energy is called damping, and
it occurs in all real mechanical systems. There are several different mechanisms that
can contribute to the overall damping of a system.
With structural damping, energy in a vibrating
structure is dissipated due to the relative motion of components at structural joints. And material damping is damping provided by
the material itself, where energy dissipates in a vibrating material due to interactions
occurring at the molecular level. To improve our spring-mass model, we can lump
the damping from all of the different sources together, and represent them by a single dampening
device called a dashpot, which is essentially a plunger that moves through a liquid-filled
cylinder. Whenever the plunger moves a force will act
to oppose its displacement, and the magnitude of this damping force is proportional to the
velocity of the displacement – the faster the plunger moves, the larger the damping
force. C is the viscous damping coefficient – it
defines the total amount of damping in the system. This model of damping is called viscous damping,
because it behaves in a similar way to viscous forces in a fluid, which are proportional
to the fluid velocity. There are other damping models, but viscous
damping is commonly used because of its simplicity.
If we include the dashpot in our spring-mass
model, the equation of motion is the same as for the undamped system, but also includes
the damping force. It's a little more difficult to solve this
equation, and the solution will depend on the amount of damping. If a system is underdamped, it will oscillate,
and the magnitude of each successive oscillation will decrease until it stops. If the damping of the system is increased
significantly, which you can think of as the dashpot being filled with a far more viscous
fluid, any oscillation will be completely suppressed by the damping. This is called an overdamped system. And a critically damped system occurs right
at the limit between these two cases – it has just enough damping to suppress vibration. Each of these cases has a different function
that defines the displacement of the system, obtained by solving the equation of motion. The ratio of the actual damping coefficient
of the system to the damping coefficient that would result in a critically damped response
is the damping ratio. Most engineering systems and structures have
a damping ratio of less than 1, so they’re underdamped.
Of course if we're modelling a real system
we need a way of figuring out which value to use for the damping coefficient. It usually has to be determined experimentally,
and one way of doing this is by measuring the displacement of the system as it oscillates. A parameter called the logarithmic decrement
can be calculated based on this test data, as the natural logarithm of the ratio of any
two successive amplitudes. The damping ratio can be calculated from the
logarithmic decrement, providing an estimate of the overall damping in the system. So far we've looked at free vibration, where
oscillation is only caused by the initial conditions – there are no externally applied
loads. But another scenario is forced vibration,
where oscillation is driven by an external force. This type of loading often occurs in rotating
machinery. A common problem with turbines and motors
occurs when a rotating component is unbalanced, meaning that its mass is unevenly distributed. This introduces a load that has a sinusoidal
component in the vertical direction, and can cause vibration.
Unbalance can occur because components were
poorly fabricated or have been distorted. But eccentric masses are sometimes added to
motors on purpose – intentionally unbalanced motors are how phones and video game controllers
are able to vibrate. We can analyse this type of forced vibration
using the spring and dashpot model, by adding a sinusoidal external load. The resulting equation of motion is similar
to the free vibration case, but is a non-homogeneous differential equation. Its solution is the sum of two functions – a
complementary solution and a particular solution. The complementary solution is the solution
to the homogeneous form of the equation, where the right hand side is equal to zero. This is just the solution to the equation
of motion for an underdamped system in free vibration that we saw earlier.
And the particular solution captures the effect
of the external loads and is given by this expression. R is the ratio of the frequency of the external
force to the natural frequency of the model. Since there's damping in the system, the complementary
solution that represents free vibration will eventually reduce to zero. At this point the motion of the system is
defined by the particular solution only. For this reason the particular solution describes
what's called the steady-state response. It has the same frequency as the forcing function,
but is offset by a certain phase angle, meaning that the response of the system lags the external
force. Something interesting happens to the steady-state
response when the frequency of the forcing load is very close to the natural frequency
of the system. R approaches 1, and so the first term in the
square root is close to zero. And if the system has very little damping
the second term will also be close to zero, which means that the displacement will become
very large.
Let's plot the normalised maximum displacement
against the frequency ratio. For an undamped system the damping ratio is
equal to zero, so when the forcing and natural frequencies match, the displacement becomes
infinite. All systems have some level of damping, so
this is just a theoretical case – here's what the response looks like for different damping
ratios. We can see this effect if we adjust the speed
of the unbalanced motor. As the frequency of the force caused by the
eccentric mass approaches the natural frequency of the system, the displacements become very
large. This happens because when the natural frequency
and forcing frequency are the same, the energy added to the system by the external force
is timed just right so that it increases the amplitude of the displacement with each cycle.
This is called resonance. Resonance can be very dangerous and needs
to be assessed carefully. It's one of the reasons it's so important
to be able to calculate the natural frequency of a system. If the natural frequency of a bridge matches
the frequency of wind loading acting on it, or of pedestrians crossing it, the results
can be catastrophic, particularly if the structure has little damping. Devices called tuned mass dampers are sometimes
installed in buildings and bridges to control the dynamic response at resonant frequencies. With a rotating eccentric mass the external
force acting on the system is a simple sine wave function.
This makes it easy to solve the equation of
motion, since we can obtain a neat closed-form solution. But if the loading is defined by a complicated
function, or is completely arbitrary, which might be the case if it's based on test data,
it might not be possible to solve the equation of motion directly, and numerical integration
methods will have to be used instead. Designing a structure to withstand seismic
events is difficult because the loading caused by an earthquake is random and can't be predicted
ahead of time. And so engineers have to use special probabilistic
techniques like the response spectrum method to design structures to withstand seismic
loads. Single degree of freedom models are really
useful, but sometimes the dynamics of a system are better modelled using multiple degrees
of freedom. Say we want to model the dynamic response
of a three storey building. If we assume that the columns between the
floors are axially rigid but can bend laterally, we can model it as a system with three degrees
of freedom, the x coordinate of each floor. Each of the masses in the model has its own
equation of motion, and if we rewrite this system of equations in matrix form we can
see it has the same familiar form as the equation of motion for an undamped single degree-of-freedom
system.
A single degree-of-freedom system has one
natural frequency and can only vibrate in one way. But since our model has three degrees of freedom
it will also have three natural frequencies, and at each natural frequency the system will
vibrate in a specific way, which is called a mode shape. Exactly how the system vibrates in practice
will depend on the initial conditions that are applied. For the three storey building the three modes
of vibration will look like this. As the number of degrees of freedom increases
it becomes necessary to use numerical techniques like the finite element method to determine
the natural frequencies and associated mode shapes. We've only covered mass-spring models in this
video, which oscillate by translating. But there are other types of vibration too,
like pendulums, which oscillate by rotating. This video is long enough already, but I've
published a short companion video that covers the motion of pendulums over on Nebula, where
we take a look at how the motion of a pendulum differs from a mass-spring system and how
to derive the equation of motion. Nebula is a streaming service built by a group
of educational creators, which means it's completely free of ads, and revenue generated
by the project is distributed directly to the creators.
It's full of curated independent content,
including Nebula originals and bonus content you won't find anywhere else. And if you're in the mood for expertly made,
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is a great way to support this channel and other creators on Nebula! And that's it for this review of vibration
and resonance. Thanks for watching!.
Let’s think about exponents with ones and zeroes. So let’s take the number 1 and let’s raise it to the eighth power.
So we’ve already seen that there’s two ways of thinking about this.
You could literally view this as taking eight 1’s and then multiplying them together.
So let’s do that.
So you have one two, three, four five, six, seven, eight 1’s and then you’re going to multiply them together And if you were to do that, you would get well 1 times.
1 is 1 times 1.
It doesn’t matter how many times, you multiply 1 by 1.
You are going to just get 1, You are just going to get 1 And you could imagine I did it eight times.
I multiplied eight 1’s, But even if this was 80 or if this was 800 or if this was 8 million.
If I just multiplied 1, if I had 8 million 1’s and I multiplied them all together, it would still be equal to 1. So 1 to any power is just going to be equal to 1 And you might say, hey what about 1 to the 0 power.
Well, we’ve already said anything to 0 power, except for 0 that’s, where we’re going to it.’s actually up for debate, But anything to the 0 power is going to be equal to 1 And just as a little bit of intuition.
Here you could literally view this as our other definition of exponentiation, which is you start with a 1, and this number says how many times you’re going to multiply that 1 times.
This number So 1 times 1 zero times is just going to be 1, And that was a little bit clearer when we did it like this, where we said 2 to the let’s say, fourth, power is equal to this was the other definition of exponentiation.
We had, which is you start with a 1, and then you multiply it by 2, four times so times, 2 times 2 times 2 times 2, which is equal to let’s see.
This is equal to 16.
So here, if you start with a 1 and then you multiply it by 1 zero times, you’re still going to have that 1 right over there And that’s.
Why anything that’s not 0 to the 1 power is going to be equal to 1? Now let’s try some other interesting scenarios.
Let’s start try some negative numbers.
So let’s take negative 1 And let’s first raise it to the 0 power. So once again, this is just going based on this definition.
This is starting with a 1 and then multiplying it by this number 0 times.
Well, that means we’re just not going to multiply it by this number.
So you’re just going to get a 1.
Let’s try negative 1.
Let’s try negative 1 to the first power.
Well, anything to the first power.
You could view this, and I like going with this definition as opposed to this one right over here.
If we were to make them consistent, if you were to make this definition consistent with this, you would say: hey let’s start with a 1 and then multiply it by 1 eight times, And you’re still going to get a 1 right over here.
But let’s do this with negative 1, So we’re going to start with a 1 and then we’re going to multiply it by negative 1 one time times negative 1, And this is of course going to be equal to negative 1. Now let’s take negative 1 and let’s take it to the second power.
We often say that we are squaring it when we take something to the second power, So negative, 1 to the second power.
Well, we could start with a 1.
We could start with a 1 and then multiply it by negative 1.
Two times multiply it by negative 1 twice And what’s this going to be equal to And once again by our old definition.
You could also just say: hey ignoring this one, because that’s not going to change the value.
We took two negative 1’s and we’re multiplying them Well negative 1 times negative 1 is 1 And I think you see a pattern forming.
Let’s take negative 1 to the third power.
What’s this going to be equal to? Well, by this definition, you start with a 1 and then you multiply it by negative 1, three times so negative 1 times negative 1 times negative 1 Or you could just think of it.
As you’re taking three negative 1’s and you’re multiplying it because this 1 doesn’t change the value, And this is going to be equal to negative 1 times. Negative 1 is positive 1 times negative 1 is negative.
1.
So you see the pattern Negative 1 to the 0 power is 1 Negative.
1 to the first power is negative 1.
Then you multiply it by negative 1.
You’re going to get positive 1.
Then you multiply it by negative 1 again to get negative 1 And the pattern you might be seeing is, if you take negative 1 to an odd power, you’re going to get negative 1 And if you take it to an even power, you’re Going to get 1 because a negative times a negative is going to be the positive And you’re going to have an even number of negatives so that you’re always going to have negative times negatives.
So this right over here.
This is even Even is going to be positive, 1, And then you could see that if you went to negative 1 to the fourth power Negative 1, the fourth power.
Well, you could start with a 1 and then multiply it by negative 1 four times. So a negative 1 times negative 1 times negative 1 times negative 1, which is just going to be equal to positive 1.
So if someone were to ask you, we already established that if someone were to take 1 to the I don’t know 1 millionth power.
This is just going to be equal to 1.
If someone told you let’s take negative 1 and raise it to the 1 millionth power.
Well, 1 million is an even number, so this is still going to be equal to positive 1.
But if you took negative 1 to the 999 999th power, this is an odd number.
