Movimiento Armónico Simple

Hello, ingenious and ingenious. Simple Harmonic Movement. In this class, I will try to explain to you what you can do about the Simple Harmonic Movement, its uses, all its applications,

mientras pueda. Este video fue solicitado por Sara Valentina. Sara Valentina es la hija de mi amigo Alex del canal Matemáticas Profe Alex. Los invito a que lo visiten. Tiene un excelente canal.

Colombian too, like me. And at the end of the lesson we will solve this problem in the movement of the piston of a car where we will find the acceleration that the piston has at the end of its career, we will find what force must be exerted at that point and the speed and kinetic energy that it has at the midpoint of its career. Let's start!

Have you ever seen the strings of a guitar? Have you ever wondered how a guitar works? What is the vibrational movement of the strings of a guitar?

When you press that string, what happens? What is the "Bai Ben, Bai Ben, Bai Ben"? A simple harmonic movement or in a sound reproducer waffler that these vibrations in simple harmonic movement form waves or a diapason?

How does a diapason produce sound? Observe the waves that are produced by that vibration, by that simple harmonic movement, or the piston of a car inside the engine, inside a cylinder, which is a movement that can be assumed as a simple harmonic movement.

Here I have a simulation of that piston in a Vaivén, Vaivén, four-stroke internal combustion engine. Let's see that it can also be applied in trolling machines like this one. Simple harmonic movement. What do I have here? A spring, a spring.

that goes and comes, I have an extreme tie ball. We are going to assume for practical effects to understand this movement that here there is no friction between the ball and the surface.

I have a constant K of the spring's elasticity. A constant that is the property of each of the elastic materials. Remember that something is elastic if when it is deformed it wants to return to its original shape. It generates a force of restitution. Here I have a point of balance.

It is assumed when the spring has not been compressed or stretched. And the extremes. How much is compressed maximum and how much is stretched maximum. This point we are going to call point x equal to zero. Remember, what is the period that is represented with the letter T? Period.

time of a cycle, time of a turn, time of a complete oscillation because in this system a complete oscillation occurs when it goes and comes, that is, we start here, it goes and comes, the time in go and come is the period for a complete oscillation, that is a complete oscillation, we can

determine it from here, when there, and it returns there, that is the period that is the time of a complete oscillation and its unit in the measurement international system is the seconds. Now, the frequency. The frequency is the number of oscillations per unit of time.

Its unit in the international measurement system is the hertz, which is also known as 1/second or second to the minus 1. It is the inverse of the period. Per unit of time in the international system is how many times it goes and comes per second, but we can also work it per minute, the rpm, revolutions per minute, okay?

or by hours, anyway, it depends on the unit you choose. In the international system, the hertz are relative to the second. And remember that the period and the frequency are inverses. If you have the period, you can determine the frequency by dividing 1 between the period or vice versa. The period is the inverse of the frequency.

The distance at any moment of the oscillatory movement of the mass with respect to the point of balance is called elongation.

If it is here, then the distance from here to here is the elongation, which can be negative if it is on the left side of what we have called as a reference point x = 0. The amplitude is the maximum elongation, it is when it reaches the farthest point of its point of balance.

And the amplitude can be positive, which we can call "a" on this side, here, or negative, which is "-a" on this other side, on the back, or left, left, yes, left, of the point that we have called point of balance. Velocities. Observe.

In a simple harmonic movement, an oscillatory movement like this one, the speed is zero at the ends.

Zero here, because it reaches there and it returns, reaches, returns, reaches and returns. And it is gaining speed because there is a force of restitution and if there is a mass and there is a force, therefore there is an acceleration. It passes through here with its maximum speed at the point of equilibrium. The speed of the oscillating mass is maximum and then it decreases, decreases, decreases until it is zero.

Increase, increase, increase, maximum, decrease, decrease, zero, increase, increase, maximum, yes, yes. Zero, maximum, zero, maximum, zero. Okay? And the acceleration is the opposite of the speed. It is zero at the point of equilibrium. Why? Because here the spring is not deformed, it is not making force. The force at this point is zero.

And I have here maximum acceleration because I have maximum force of restitution and here maximum force of restitution. Therefore, the accelerations are maximum at the extremes and zero at the point of equilibrium. And of course, obey the Hooke law. Do you remember the Hooke law? That the force of restitution of the spring is directly proportional to the deformation of the

and observe that if the elongation is positive the force is negative therefore this sign and this is the proportionality constant of the Hooke law and it is the K constant of the spring the constant of restitution and the Hooke law is fulfilled when the elongation is negative the force is positive look that they are vectors with

opposite senses, therefore, there is this minus, that sense, the force has a sense opposite to the elongation, to the elongation vector. Attention to this, very important, if in a periodic movement, periodic movement, periodic movement, vibrational like these, oscillatory, the force of restitution F

is directly proportional to its displacement x, that is, if the Hooke law is fulfilled, then this movement is simple harmonic. When a periodic oscillatory movement, vibrator is simple harmonic, when the force is directly proportional to the elongation, to the displacement. If we know

According to Newton's second law, this force is equal to the mass product by its acceleration and we substitute this force here, let's put it here, instead of F we are going to write mass by acceleration, then this mass that is multiplying goes down to divide and I can determine the acceleration of this mass in that spring as minus the constant of the spring divided by the mass by

the elongation x. Energy in a simple harmonic movement. Remember that energy is the capacity to perform a work and here there is work because there is speed and because there is deformation of a spring and remember what is the energy of speed?

Where does this system have the maximum speed? This system has its maximum speed at the point of equilibrium and there we will have all its kinetic energy. Do you remember the mathematical relation that serves to determine the kinetic energy?

Very well, a half of the mass by the velocity that it carries to the square and at this point the velocity is maximum and at the ends here when it is completely compressed at the point of its negative amplitude

I have elastic potential energy and it can be calculated as a mean of the elastic constant of the spring by the deformation to the square. Therefore, all this kinetic energy is transformed into elastic potential energy. And the same here.

When it stretches and reaches the farthest point where I have its positive amplitude, also all this kinetic energy has transformed into elastic potential energy, law of mechanical energy conservation. If there is no friction, if there are no energy losses due to non-conservative forces, then remember that we are ignoring friction.

and all this energy is transformed into this one, a half of the elastic constant by the square amplitude. And what happens if we are between the point of equilibrium and the extreme? At a point where I have an elongation x, there I have the two energies. Why? Because I have a mass with velocity and a mass that has deformed a spring.

Therefore, I have kinetic energy plus elastic potential energy, one half of the mass by the velocity that the mass carries at this point squared, plus one half of the elastic constant by the elongation squared.

Here there is no gravitational potential energy in this example because it moves horizontally, it does not change in height. If I have a mass that is oscillating vertically, then yes, you have to take into account the gravitational potential energy. In this case, no. And these two added give me the total energy of the system in this movement.

And notice that the sum of kinetic energy plus potential energy is equal to the elastic potential energy at the end, with all the amplitude. That is the energy. Now, what are we going to do? Look, what can I do with that equation? There I have an algebraic equation that represents an equality in the total energy of the system.

What can we cancel here? Can we cancel the mass?

Not because the mass is not in all the terms. We can cancel x, neither k. k is here, it is here, but it is not here. We can cancel a half, a half and a half, yes. We are going to cancel them, the three. Tin tin tin. What do I have left? Mass by the square of the speed plus k by x squared equals k by the square of the amplitude. The intention that I have in this case with this equality

is totally valid in a simple harmonic movement is to clear the speed a way to calculate the speed of an oscillating system having its mass, the elastic constant, its elongation and its amplitude. There, take a sheet of paper

Your notebook and solve it. Clear the velocity in that equation. How would you do it? I have an equality. I want to clear the velocity. What is the first thing I remove from the mass? No, no, no. First I remove this term to leave this term alone. kx squared that is adding goes to subtract k squared. And observe mathematically here what I can do. I have k and k.

is a common factor. Let's factor. Remember, in the case of factorization, if you factor, you get k as a squared minus x squared, that is, the elastic constant as a factor of the amplitude, which is the maximum elongation,

squared minus the elongation squared. And here, if I want to clear the velocity, this mass that is multiplying goes to divide and I only have the inverse operation of elevating to the square, which is to determine the square root.

I have the magnitude of the speed or speed of this oscillating mass, this velocity, and it is calculated as the square root of the elastic constant multiplied by the square amplitude minus the square elongation and this between the mass, its root. Ready? Now, I found this scheme on the internet

about the relationship between a uniform circular movement and the simple harmonic movement between an MSU and a MAS. This red ball is in a uniform circular movement because it has the same angular displacements at the same times, that is, its angular velocity is constant and the projection

of this red ball. If you hypothetically put a sphere in space and you illuminate with a lamp and see the projection of the shadow of this sphere on a wall,

you will see an oscillating shadow and that describes a simple harmonic movement. Therefore, there is a relationship between a circular movement and a uniform and a simple harmonic movement. And there it is, clarity. We are going to use this same scheme, but in our system. I put the ball in circular motion in a radius of radius A, where A is the amplitude,

of the oscillating system and if now I am at this point it is because this ball is at this point an angle has been displaced theta or tita this angle and I have that this is the radius of the circumference which is the same amplitude now the projection of the vector a on the x-axis gives me

the elongation x, I have the elongation vector, this is the speed of this ball in uniform circular motion, I have the angular velocity or the angular frequency of this simple harmonic movement, remember the difference that there is a difference between the frequency

which is the number of oscillations per unit of time and the angular frequency, which is symbolized with the Greek letter "ω", angular velocity or angular frequency, "ω", and is calculated as 2π radians, that is, a full turn, by the frequency, number of turns per unit of time or number of oscillations per unit of time. So, the angular frequency of this oscillating system

is calculated as 2 pi by the frequency and its unit in the international measurement system are the radians over the second. 2 pi radians, whole turn, over the second. And since the frequency is the inverse of the period, you can also replace, replace this frequency in this product, you pass it to the denominator as period.

And you have another relation, also very useful and very important, in the uniform circular movement and in the simple harmonic movement, as the angular frequency is 2π/period. From this point to this point, this little ball describes an arc S and uses a time t. Now, do you remember how this tangential or linear velocity is calculated in a uniform circular movement?

You can find a very good explanation of Uniform Circular Movement here on my channel. Search, write, Uniform Circular Movement MSU, Professor Sergio Llanos. Do you remember then the mathematical relation to calculate the tangential or linear velocity in a Uniform Circular Movement? Here it is.

V is equal to the angular velocity omega times the radius. And since the radius is A, that is, the amplitude of this system, then this tangential or linear velocity is the angular frequency times the amplitude. And also remember what is the fundamental definition of velocity. In this case, the speed is the arc

divided by the time, that is, the trajectory, the distance traveled between time and we can substitute it here as this speed is the arc, the arc, this, over time. And now we are going to make an exchange between

and T algebraically this T by transposition of terms that is dividing goes to multiply and this A that is multiplying goes to divide we can make the exchange let's do that algebraic exchange I have left this way and now remember what is arc over radius the arc over the radius

The arc over the radius is the angle Z. You can calculate this angle in radians

This angle in radians is the arc divided by the radius. This is the angle theta. And I have that this angle is equal to the angular frequency over time. And we are going to substitute it here. This angle is omega t. Now with all these elements we are going to calculate, determine the

elongation in a simple harmonic movement. The elongation, like this vector, that marks the displacement of the harmonic oscillator in its position from the point of balance to where it is. Remember, trigonometrically, observe that here I have a rectangular triangle

Angle, cathode, hypotenuse. What trigonometric reason articulates this cathode, which is adjacent to the angle, and the hypotenuse? Very well, the cosine. Therefore, if the cosine of omega t is the adjacent cathode of x over hypotenuse A,

I have that x is a by the cosine of omega t. This is my relation to determine the elongation in a simple harmonic movement. Let's leave it here. Let's go with the speed in a simple harmonic movement and from this speed we will find the horizontal projection of this speed. Let's put this auxiliary horizontal line.

and this vector is the projection of the velocity vector, let's call it Vx and this Vx is the velocity, this velocity of this harmonic oscillator. And how do we calculate it? For this we use some geometric tricks, trigonometric. Angle, this angle omega t in this triangle, where does it go? How does it go? How do we articulate it? Observe

that if the angle omega t is in these two green vectors, its perpendiculars, this red perpendicular of this vector and this red perpendicular of this vector, both perpendiculars form the angle omega t and the angle omega t we can place it at this point here. And since this is omega t, the angle

the velocity in X is the opposite of omega t and being the opposite of omega t in that rectangle, observe that this velocity has a sense opposite to this elongation. As this elongation is considered positive, this velocity points in the opposite direction to the left, we will have it negative.

and is equal to the velocity by the sine of the angle omega t and remember that the velocity is the angular frequency by the amplitude and we already have the relationship to determine the velocity of this harmonic oscillator let's leave it here

Elongation, velocity. Notice that this is a function, the elongation as a function of time, because the amplitude is constant, the angular frequency is constant, because it is a simple harmonic movement, and the time is variable. I have a trigonometric function, the elongation as a cosine function of time, and the velocity as a sine function of time, a negative sine.

Okay? What do we need? Of course, acceleration in a simple harmonic movement. And let's start with what is the acceleration in a uniform circular movement? What acceleration is there in a circular movement? Of course, the centripetal acceleration, which is what makes the trajectory of a particle, of an object, circular.

Ready? So, centripetal acceleration. This centripetal acceleration will obviously also have its projection on the horizontal axis, this triangle, rectangle, and this is the acceleration of this object that is oscillating.

and this is the angle omega t. Why is this the angle omega t? Because I have two parallels, a transverse and these two angles are internal and as this is the angle omega t and this is the adjacent cathode, it becomes cosine again. But the direction is the same, but the sense is opposite, opposite. It is also negative and it is the centripetal acceleration, this centripetal acceleration,

by the cosine of the angle omega t. And do you remember how to calculate a circular movement uniformly of the centripetal acceleration? Math relation for the centripetal acceleration? Of course, the square velocity over the radius. This centripetal acceleration, the square velocity over the radius, and in this case the radius is the amplitude. And remember that the velocity

is the angular frequency by the radius, which is "a", and this squared. Algebraic property of a product squared, we can distribute this power for both, both for the angular frequency and for the amplitude.

and we can cancel amplitude here and here and I have the angular velocity squared by the amplitude that is the centripetal acceleration that we are going to substitute here and we already have the relationship

the acceleration as a function of time, a negative cosine function. And now what can you observe here? Look that the amplitude by the cosine of omega t, what is this equal to? This here, this is equal to the elongation, look, the elongation. Therefore, I can rewrite the acceleration of an oscillating system

as minus the angular frequency squared by the elongation. Note that we have worked this oscillating system with the initial angle, with this displacement, with this time, this angular displacement in time.

from here to here. But what happens, let's put these equations over there, what happens if we don't start at the end, but a little further over there, a little further over there, a little further over there? First, an angle Φ, let's call this angle Φ, Greek letter Φ. If we start at Φ, this is going to be our ωt, this is going to be our displacement, so we can calculate the elongation

the velocity and the acceleration increasing this angular deface in the angle, that is, omega t plus phi, omega t plus phi. If we start at this point, that is, at the end, then phi is zero and that's it. We have that the elongation of an oscillating system at that point is its amplitude by the cosine of its angular frequency over time plus the phi deface, that its velocity

which is in the opposite direction to the elongation, can also be calculated as minus the angular frequency by the amplitude by the sine of the angular frequency by the time plus the phase-diff and that the acceleration in a simple harmonic movement is minus the angular frequency squared by the amplitude by the cosine of the angular frequency by the time plus the phase-diff.

and the same thing we said before. I want to show you a simulator with an oscillating system that shows you how vectors, velocity and acceleration can behave in a simple harmonic movement. The acceleration vector is the red vector and the velocity vector is the blue vector. I want you to observe. When it passes through the balance point

When it passes through the point of equilibrium, observe that the acceleration, as we had said before, is zero. Here. And it is zero. Why? Because the spring is not compressed or stretched. The force is zero, the acceleration is zero. But I have all its maximum speed. Look that the vector here is changing. Okay? Ready. Very good. At the ends. At the ends, look that the speed is zero. Here it is in this direction. Zero.

zero and at the ends the acceleration is maximum because I have the maximum stretching or compression of the spring and it is showing me the acceleration based on the force of restitution. Look that the spring goes over there, pushes, then the acceleration goes in the same direction. Here the acceleration vector changes direction, that is, the force vector. And I wanted to bring you

this relationship. Look, what are we going to identify here? I have three circumferences in a circular movement, I have the elongation, the speed and the acceleration. Look that this elongation is the projection, this is the simple harmonic movement, but I'm going to stop it

when we are here. Stay still there. And we are going to identify the graphs of the elongation, speed and acceleration. Why did I put them in this circular system? Because they are trigonometric functions and I can graph a trigonometric function from a trigonometric circle of the radius depending on this coefficient. I have

The cosine, look at the graph of the cosine, it is the green one, which is the amplitude and I have a cosine in its first period. It comes back and repeats itself, but in the first period I have a positive amplitude cosine. The velocity,

is a negative sine. Look at the speed, it's the one in red. I have a graph of the negative sine. At the point of equilibrium where the oscillating system is in equilibrium, I have its maximum speed and it is in equilibrium again and I have its maximum speed and acceleration zero. Look at the acceleration is zero at the point of equilibrium and I have maximum speed. And

the maximum speed at the angular frequency by the amplitude. I have maximum speed at angular frequency by positive amplitude and maximum speed at angular frequency by negative amplitude. Here I have the two speeds. And with the acceleration, the acceleration is a negative cosine. This negative cosine where I have its maximum acceleration at the points where it is zero

Here in these points I have maximum accelerations in omega squared by a and in minus omega squared by a. Therefore, we can determine that the maximum elongation is given, this maximum elongation, this maximum x is given in a, in the amplitude.

that the speed is maximum, the red is maximum in the angular frequency by the amplitude and that the acceleration is maximum, where? where? where? where? in the angular velocity squared or the angular frequency squared by the amplitude in an oscillating system, there are the graphs, analyze them, interesting, right?

I was missing to say, to affirm that all this that we have just done is with phi equal to zero. Because we are starting at this point. Now we are going to determine some fundamental mathematical relations important in a circular movement to solve the problem at the end of the class.

I have, as we had already manifested, that the acceleration is minus the angular frequency squared by the elongation. I have the Hooke law where the force is equal to minus the constant of the elastic by x and according to the definition of force, force is equal to mass by the acceleration and we can substitute this acceleration for minus omega squared x.

and by doing this substitution, what can I cancel here? I can cancel this negative with this negative and this elongation with this elongation because they are products and it is an equality and I have that the elastic constant is the angular frequency squared by the mass. If I am going to clear the angular frequency, this mass that is multiplying by transposition of terms goes to divide

and this square raised, I get square root and I can find the angular frequency of an oscillating system like the square root of the elastic constant divided by its mass. Ready? Important relationship. Also, remember that the angular frequency is 2π over the period. If we alternate period with angular frequency, tin-tum,

We are going to find the period in an oscillating system, in a spring like the one we are working with. We substitute this as root of k over m and we organize the law of the ear, this is 2 pi, or the law of extremes and halves, root of this m raised to multiply over k and it is another important relation to calculate or determine the period in a spring.

This is the relation to find the period in a resort. Let's group everything we have worked so far in this table. Remember that the notes of this class will be down here in the video description and you will be able to find this well organized. This is what we have worked so far. Now let's solve a simple harmonic movement problem. This problem

is taken from the book of university physics by Siersemanski and I will solve it here these three dots A, B and C that indicates the movement of the piston of a motor car is approximately a simple harmonic movement if the piston race the double the amplitude is 0.1 meters and the motor works 3500 revolutions per minute

What acceleration does the piston have at the end of its race? Here I have the race, acceleration here. Part B: If the piston has a mass of 0.45 kilograms, as in the book we were working on the decimal point indicator, I left it as a point. Remember that we have to specify if the decimal point indicator is a point or a comma. In this class, in this video, the decimal point indicator will be a point.

0.45 kg, what net force should be exerted on it at this point, there at the end, and what speed and kinetic energy does the piston have at the midpoint of its career? Over here. Let's go with the first part, the A. If the piston's career, which we are going to assume from here to here, is twice the amplitude, the amplitude goes from the balance point, I have drawn this guideline line,

this is the amplitude, then the 0.1 meter race is twice the amplitude. If this race is twice the amplitude and it is 0.1 meters, the amplitude is then half the race and 0.1 meters divided by 2 gives me 0.05 meters. I already have my amplitude. Now,

What is this that they give you here? That the motor works at 3500 rpm or revolutions per minute. What are they giving you there? The frequency.

and that frequency is 3500 revolutions per minute. We are going to convert this RPM to the international unit system, that is, to the Hertz. Let's go to the Hertz, 3500 revolutions per minute, let's pass these minutes to seconds. Let's multiply it by, what goes in the numerator? Minutes or seconds?

Since we are going to cancel minutes with minutes, one minute here is 60 seconds. We cancel minutes with minutes, 3500 times 1 over 60 gives me 58.33 hertz. 33 periode, but we are going to keep these 2, 3. And we are going to work with roundabouts, however, I left all the calculations in the calculator to continue using them later.

I already have my frequency of 58.33 Hz. We have the amplitude and the frequency in this piston. Remember that in every car, in every motorcycle, it has an internal four-stroke combustion engine.

that uses as fuel, in this case, gasoline. Here I have in a piston, four valves, two for intake and two for exhaust. This is the nozzle that generates the spark, this is the piston inside the cylinder, this is the connecting rod and this is the cork.

in a four-stroke internal combustion engine that what it does is transform the chemical energy of gasoline into thermal energy, into a thermodynamic expansion, into the explosion of gasoline,

pushes the piston, generates a force, does a work on the piston and it transforms into mechanical energy, which is what makes a car move. It's like the bicycle you have and what your leg does. What it does is transform the chemical energy of food into mechanical energy, well, into a lot of energy.

in thermal energy because our body must work hotter than the medium, at approximately 36 degrees Celsius. Our body needs electrical energy

Why? Because our whole neuronal system works with electricity. Our heart, which is a hydraulic pump, works with electrical energy. The myocardium is compressed, contracted with electrical impulses. And where do we get that energy? From food. That's why you have to feed yourself very well.

But let's continue with our problem in our four-stroke internal combustion engine. You may ask, why four-stroke? The four-stroke are: admission race, we have here the admission race here, suction, compress, burst and take out, enters blue gasoline,

Here, compresses, explodes, the spark makes that mixture of air and fuel that comes from here, from the injector, look that air enters here and fuel here, this is the air and fuel injector. So, air and fuel, compresses, explodes, here boom, work race, of the four, it is the only race that does work and then these valves are opened and the smoke is taken out as a combustion product by the exhaust.

There are the four times in a motor. These two valves are opened. Admission, compression, work and escape. But let's go back to where we were. What acceleration does the piston have at the end of its career? So, which of all these relationships helps us to determine the acceleration of the piston at the end of its career? With what we have, what do we have? The amplitude and the frequency. So I have acceleration here.

but we don't have time. We have acceleration here. Ah! But they tell us at the end. At the end, then, it is its maximum acceleration. Here it is. Yes, you see? Look at it. Here it is. So, this is the acceleration.

Do I have the angular frequency? No, but I can calculate it. Why? Because I have the frequency and with the frequency multiplied by 2π I have this angular frequency, I have the amplitude, there it is. With these two we are going to use them, here they are. Let's start calculating the angular frequency, 2π radians multiplied by the frequency which is 58.33 Hz.

2 pi divided by 58.33 gives me 366.5 radians per second. Hertz is 1 over second, radians per second. Ready? And now the acceleration.

The acceleration, remember that it is at this end, therefore we will have maximum acceleration. Here at this point I have this amplitude that we already have of -0.05 meters and we substitute the angular frequency that we already calculated, squared by the amplitude that we also have.

less by less is more and we do this operation in the calculator and it gives us 6716.81 meters per second squared. Ready? That is point A. Let's go to point B. We are leaving here the results obtained and the mathematical relations used. If the piston has a mass of

of 0.45 kilograms this piston has that mass, what force should be exerted on it at that point? So if I have this mass of 45 kilograms, what force, what force is exerted? Remember that force is simply mass per acceleration.

I have the mass, I have the acceleration, I replace, I substitute, I do the product and I have 3022.57 newtons or 3.022 kilonewtons. That was point B, that's it. Let's go with point C, the last point where they ask me the speed and kinetic energy at the balance point, here in the middle of the race.

this kinetic energy and this velocity. And what do we do then? The maximum velocity, because at the point of balance the maximum velocity is the angular frequency by the amplitude. And I have both. I have the angular frequency, I have the amplitude, I substitute, ping, pam, calculator, and I have 18.33 meters per second. That's it. That is its velocity, both going up and down. And the kinetic energy?

Also, kinetic energy is a mass medium at this square velocity. I have mass, I have velocity, and I have a medium of 0.45 kilograms, for its square velocity, which gives me 75.56 joules. Why joules? Because it is the unit of energy, because kilogram per square meter over the second square is joules. And that's it.

We have solved our problem. I am Professor Sergio Llanos, Mechanical Engineer at the University of El Valle in Cali, Colombia. If you liked this class, give it a like, subscribe to my channel, activate the bell. Remember that all the notes of this class will be downloadable down here in the description of the video. Have a great day.