Physics 19 Mechanical Waves (2 of 21) Velocity on a String (Deriving Equation)

[@bushidobrownbaby662]

This guy just went super saiyan deriving that equation

[@EternusVia]

That was honestly a phenomenal video - perfectly explained in clear and concise terms.

[@roshanaab116]

Your videos are absolutely great! Keep up the amazing work. Your initiative and effort has definitely impacted many students across the globe who wish to learn more.

[@ShuusakuSama]

despite I know the end result, I did not know that's how you derive them . . . great job, I get why the other guys in the comments said he went super seiyan mode

[@ahmedal-ebrashy3691]

One of the harder videos and I am honored to have watched this...This was so phenomenal :D

[@elmerjones8075]

This is a great alternate proof to how the textbooks portray relating Vx to centripetal acceleration.

[@isaiasramirez2909]

my physics teacher is amazing but you are next level amazing at explaining things

[@SimantaKumerPundori]

- 00:00 The velocity of a wave on a string depends on tension and mass per unit length of the string. - 02:13 The velocity in the Y direction can be expressed in terms of velocity in the X direction as the velocity of the wave times the tangent of theta. - 04:26 The tension times the sine of theta DT is equal to V in the Y direction DM, which can be simplified to T times DT is equal to V times DM. - 06:38 The velocity of a wave on a string is equal to the square root of the tension of the string divided by the mass per unit length. - 08:51 The velocity of waves on a string can be found using a derived equation involving impulse and mass per unit length.

[@MichelvanBiezen]

Thanks!

[@at8736]

Great lecture. You remind me of mr. noodle from elmo's world.

[@jgvlogs5358]

Jalal , 😂😂😂 he does!!

[@computerlover9290]

But in 4:18 velocity in the y direction isn't always constant! It is positive when going up and negative when going down! Can you please clarify that for me ?pleaseeeee!

[@oguz.ozer2004]

Do u have clarification now? Kuz I need too

[@sarunyuyu4268]

Thank you,Sir for wave equation

[@zakirhussain-js9ku]

The wave on the string & EM wave appear similar except that string has mass whereas EM wave is massless. Above formula will yield infinite speed if we enter 0 mass in the formula. Therefore light has to have some mass to yield finite speed.

[@MichelvanBiezen]

Waves on a string are not a string, they are two different entities. Waves represent energy and travel along the string which is the medium

[@zakirhussain-js9ku]

​​​​@@MichelvanBiezenA sine wave on a string made up of electrons should constitute mechanical wave since electrons have mass & coulomb force. In a conductor it represents alternating current which travels at light speed. Why formula for string should not be applicable to light which like current also travels at light speed.

[@MichelvanBiezen]

"A sine wave on a string made up of electrons " --> that is impossible

[@zakirhussain-js9ku]

​​​​​@@MichelvanBiezenA sine wave on string made up of electrons is just a thought & such a string may be impossible, but how do we know for sure such a string carrying transverse wave for ac & longitudinal wave for dc does not exist in a current carrying conductor. Electrons move very slowly, only a wave could travel at light speed in a conductor. If we put 9.1×10-³¹kg mass of electron & 2.817×10-¹⁵m its diameter in the formula for speed of mechanical wave we get, 299,796,881 m/s speed of wave on electron string with electrons stacked back to back. Coulomb force = 9×10⁹(1.6×10-¹⁹)²/(2.817)²×(10-¹⁵)²=29.0341276.N Mass of electrons per meter length = 9.1×10-³¹/2.817×10-¹⁵. Dividing Coulomb force by electron mass per meter we get. (8.987817×10¹⁶)¹/² = 299,796,881m/s. Given uncertainties in mass, diameter of electron & calculations above result may not be valid, but wave on a string made up of electrons & above results appear quite fascinating.

[@zakirhussain-js9ku]

​@@MichelvanBiezenA sine wave on a string made up of electrons may be impossible. During current flow electrons move very slowly but signal moves at light speed I thought may be it is wave which travels through electrons. If we enter 9.1×10-³¹kg mass of electron & 2.817×10-¹⁵ m dia. of electron in the formula for speed of mechanical wave we get 299,796,881m/s speed on a string made up electrons. Coulomb force = 9×10⁹ (1.6×10‐¹⁹)²/ (2.817)²×(10-¹⁵)²= 29.0341276N. Mass of 1meter long string of electrons= 9.1×10-³¹/ 2.817 ×10‐¹⁵ Dividing Coulomb force by mass of electrons per meter we get (8.987817×10¹⁶)¹/² = 299,796,881 m/s. Given uncertainties relating to mass, diameter & applicability of formula above result may not be valid, but electron string & speed result look interesting to me.

[@Supermario0727]

If you have a function for displacement, can you calculate the velocity of the wave by finding the derivative of the displacement function?

[@MichelvanBiezen]

+John Stuart Note that there are two types of displacement. In the y-direction you have the displacement of sections of the string perpendicular to the motion of the wave (energy). In the y-direction you have the displacement of the wave itself which carries the energy along the string.

[@taufiqurrohman8987]

thank you very much sir

[@MichelvanBiezen]

Most welcome

[@willpaisley7828]

Great Video, really helpful for my revision. Thanks!

[@richardbruno201]

You are the GOAT

[@MichelvanBiezen]

Thank you. Glad you found our videos! 🙂

[@nitishrishi807]

good explanation liked this sir

[@hak9786]

Can you please tell me how the force on the string at 3:07 is in the upper direction? See, the tension is pulling the string in both the horizontal directions. What if the wave at that point is a trough? Then will the force be downwards? Thanks!

[@MichelvanBiezen]

When the string at any point is at an angle (not horizontal), then the tension in the string will have 2 components. A vertical component and a horizontal component. The vertical component will pull the string upward in this example.

[@name-qb8ec]

@@MichelvanBiezen is horizontal component of tension balanced? If so how?

[@alwaysdisputin9930]

So trying to understand this intuitively, if you have a heavier string, does the wave move slower? Yes. v = √ T/μ & μ is mass per length Why does it move slower? Because if we zoom in to a small area at the front of the wave, the upwards momentum vᵧ dm = the tension in the string, T times dy ̅d̅ ̅x ̅ where dy = the amount the wave moves up in a small amount of time, dt & dx = the amount the wave moves right in a small amount of time, dt I'm not really getting an intuitive feeling for what's happening. I'm kind of looking for some sort of explanation that you could put in a video & it'd make sense to people. if I can't do that then It's just formulae that I don't really understand. It's not enough that the derivation works mathematically. I need to learn the derivation inside & out & think about this a lot in intuive terms & maybe I'll reach a deeper understanding

[@MichelvanBiezen]

The same amount of force applied to a more massive object will cause the acceleration to be smaller. (F = ma) Also note that the oscillations of a pendulum will slow down with a more massive object.

[@curtpiazza1688]

Excellent!

[@MichelvanBiezen]

Thank you. 🙂

[@michaellewis7861]

Tcos(theta)*dt=dm*v.... that’s all the needed to be done.. cos(theta)=1, Tdt=dm*v, or =(mass density)dx*v T=mass density*v^2 sqrt(T/mass density)=v... isn’t the derivation show a bit unnecessarily convoluted by doing it in terms of vy?

[@ahmedal-ebrashy3691]

so all waves have velocities in the x direction and y direction and z direction and all are equal in magnitude?

[@MichelvanBiezen]

Not for a wave on a string, the velocity there would only be in one dimension.

[@MohammadYaseen-oz6lo]

I don't understand how and why the momentum matters here ?

[@brianronan1905]

Is there a way to calculate Fy given tension, mass, and length? Would Fy represent the amount of horizontal force in a given dx length of vibrating string? Can theta be measured for an arbitrary wave or "pluck" of a string? Things I wonder

[@AllouhaX]

what does {T} stand for ?

[@MichelvanBiezen]

T is tension

[@eien7228]

F

[@arunsharma4221]

Thanks Sir 👍

[@HajarZhiri-t4u]

excellent

[@MichelvanBiezen]

Thanks!

[@hak9786]

Also at 3:38 you say that f=t sin there. And you've assumed that the string is not massless. So it should be f=t sin theta + mg. So it'll change almost everything

[@MichelvanBiezen]

The force in the vertical direction is equal to T sin(theta). The mass (and weight) of the string for that section is negligible compared to the magnitude of the tension and can thus be ignored in the vertical component of the force calculation.

[@hak9786]

@@MichelvanBiezen thanks for your reply!

[@Avilapardede]

Why is the v_y constant? I have just moved from simple harmonic oscillations to waves, and I thought that the velocity in y direction changes. Do the y-direction velocity of waves and oscillations work differently? Thank you for the video!

[@MichelvanBiezen]

Vy is not constant, but constantly changing in the same way as the velocity of an object that experiences simple harmonic motion.

[@Avilapardede]

Thanks for the reply! When explaining the change in momentum/impulse, you said that the velocity to the right is constant, and so is the velocity of the upward direction. Does vx affects vy? How do they affect each other? I'm sorry if I misunderstand something.

[@parmveercharak2179]

sir i have little doubt . as change in momentum = change in momentum= change in velocity * mass. but you took it as change in momentum = change in mass * velocity. how it is possible bcoz mass is conserved .

[@oguz.ozer2004]

i have same doubt. Can u find any answer for this

[@MichelvanBiezen]

Here the "dm" is not the change in mass, but a small segment in mass (this is the calculus method)

[@oguz.ozer2004]

But you said Vy is constant and m changes. I think m should be constant and Vy should change, kuz Vy is not equal to V. @@MichelvanBiezen

[@imen2015]

How did you cancel sin and tan on both sides?????

[@MichelvanBiezen]

For small angles, the sin is equal to the tan.

[@vedantsinghal5637]

sir, i did understand how trignometric functions are negligible as when i tried to solve , i was getting [1/cos(θ)] in RHS

[@imen2015]

Thank you very much

[@onehourwithclassics]

Thtank You Sir..

[@historyisthebest5831]

Hello sir. I just finished learning the topic of Thermodynamics, though I will review it frequently, what topic do you think I should learn next?

[@MichelvanBiezen]

That is actually up to you. We tend the gravitate to the subjects we like best

[@erwinjedracho640]

neat proof

[@AlexandrBorschchev]

can you derive without calculus?

[@MichelvanBiezen]

Most derivations like that require calculus, including this one.

[@ofofo-k3o]

Why is velocity in the upward direction constant?

[@MichelvanBiezen]

It is not constant. It is constantly changing. (Like simple harmonic motion)

[@ofofo-k3o]

@@MichelvanBiezen but then aren’t we supposed to use change in v instead of just v for the impulse. So it’d be I=dvydm?

[@MichelvanBiezen]

Note that the Vy is related to V through sin(theta) which for small angles is the same as tan(theta). Also the ratio between Vy and V is a constant and dependent on the Tension. And that is how Vy is related to V

[@jycwong112]

3:25 Fy is from the tension Do we need to worry about the other tension from the right ?

[@MichelvanBiezen]

+jycwong112 There is only 1 tension, namely the tension in the string. It does seem odd and it looks like there is tension pulling to the left and there is tension pulling to the right, but it is the same tension. To make it seem easier, always consider the tension with respect to a point or with respect to an object. With respect to a point on the string, the string pulls to the left with tension T and the string pulls to the right with tension T and the two cancel and thus there is no acceleration.

[@MichelvanBiezen]

+jycwong112 Fy is the component of the tension in the y-direction.

[@danielm9463]

+Michel van Biezen So there is tension in both directions, and we assume they are equal. The reason for the upward acceleration is that the left-side tension is now directed up along an angle. And so there is a net/ unbalanced force on the y axis (namely the y-component of the left-side angled tension), and this unbalanced vertical force produces the change in vertical momentum. When using FnetΔt = (Δm)(vy) in the y direction, we can ignore the two x-direction forces, namely: (a) the rightward tension and (b) the x-component of the other tension which is angled leftward and upward. But if we assume the tension is constant as the wave travels across the length dx, then during this interval dt, wouldn't the string slide to the right slightly because (a) the rightward tension is bigger than (b) the x-component of the leftward tension? We can imagine tension as resulting from attraction to neighboring particles in the string due to slight stretching when taut. The left edge of the small bit of rope is attracted leftward with a force Tcosθ, and the right edge of the small bit of rope is attracted rightward with a force T. Doesn't this produce an unbalanced rightward force equal to T - Tcosθ? Also, we know from simple harmonic motion that the rope particles have different speeds when at different vertical positions, yet we're using a single speed vy to describe the entire length of rope dL. We also know that dy is only the vertical displacement for the leftmost edge of the small bit of rope, and the vertical displacement for the rightmost edge of the small bit of rope is 0. What is our assumption about dt? Are we assuming dt is so small that (a) the entire rope has a constant vertical speed and (b) the entire rope has moved up by the same amount dy? This seems wrong, since we rely on the presence of an angle in the derivation. Additionally, if the vertical speed is increasing, how can we assume a constant vy when writing the change in vertical momentum Δmvy? Why isn't it ΔmΔvy?

[@danielm9463]

+Michel van Biezen It is true that we can assume constant vertical speed. The SHM equation would be vy = ω sqrt(amplitude^2 - dy^2). Since we can assume dy

[@ahmedmostafa-bn8tn]

honestly, I can't understand anything, all I have just thought about while listening was how poor I am in math and physics and the books and cyclopedias and books of math and physics in black and white :( any suggestions?

[@MichelvanBiezen]

What is your background? Age, level of education, and what are you trying to study? Usually we start at the beginning and slowly work our way through the material trying to understand it one step at a time. This channel is organized to present math and physics in a very systematic way so that you can learn everything starting from whatever level you are at.

[@ahmedmostafa-bn8tn]

@@MichelvanBiezen thanks for the reply 💚 I really appreciate your great work and efforts ❤️. So the case is that I thought I would study physics easily even if I didn't learn or strongly grasped the diffirentiation rules... But hopefully I started learn from where I stopped and I would be able to continue your list in the oscillating motion and others ^_^ Also I am in grade 12 Thanks for the great efforts that affect millions of students ❤️

[@MichelvanBiezen]

It is not necessary that you understand every aspect and every video as you try to learn physics. Even if you understand just 80 or 90 % of the videos, you will still have learned a lot and as you continue to study physics you will understand it better and better. (It is not an all or nothing things). This particular video is a bit more challenging, but you can skip it for now.

[@sayantantalukdar9883]

what about gravitational force on dm which is acting in Y direction?

[@MichelvanBiezen]

Typically the tension in the string is far greater than the gravitational forces, and they are therefore ignore in a problem like this.

[@Thepankajsinha123]

Sir , but dL should be √(dx2 +dy2) not dx please clarify.....

[@MichelvanBiezen]

Pankaj Note that the dL is used to calculate the mass of a small section of string. Therefore dm = mu * dx

[@MichelvanBiezen]

Pankaj Sinha In this case dL simply represents a small piece of the string

[@danielm9463]

+Pankaj Sinha It's true that the small bit of mass dm is directed along an angle at the moment t + dt. So it would seem that we must use μ dL (not μ dx) since the mass dm is not directed along the x axis. Because the angle θ is small, we can assume dx = dL. This follows from the small angle approximation: sinθ = tanθ (1) sinθ = dy/dL (2) tanθ = dy/dx (3) substituting (2) and (3) into (1) gives: dy/dL = dy/dx and from this we see that dL = dx.

[@avdeshkumar8500]

how to use calculus in a right way to get the answer

[@MichelvanBiezen]

+AVDESH KUMAR Calculus is something that takes a long time to learn, and there are different aspect of calculus that need to be understood. The best thing to do is to take a few courses in calculus. To get a feel for it, you can watch the calculus videos on this channel. Those will probably answer a lot of your questions.

[@avdeshkumar8500]

+Michel van Biezen but how to know the assumptions that you assume

[@MichelvanBiezen]

+AVDESH KUMAR That is something that comes with experience. When you look at several different systems and you begin to compare them, you'll begin to see common trends. That is why we make these videos, so you can have examples.

[@reynalyngeron9278]

okinnana nagrigat la kitden

[@MichelvanBiezen]

Not sure if we understood your comment correctly.

[@aishawaheed7377]

Can you plwase take a side i have to take a screen short (lol)😂

[@9_kicksfooty]

Too fast

[@MichelvanBiezen]

Thank you for the feedback

[@9_kicksfooty]

@@MichelvanBiezen I did understand later though so thank yiu