|Ch 01: Units & Vectors||2hrs & 22mins||0% complete||WorksheetStart|
|Ch 02: 1D Motion (Kinematics)||3hrs & 11mins||0% complete||WorksheetStart|
|Ch 03: 2D Motion (Projectile Motion)||3hrs & 8mins||0% complete||WorksheetStart|
|Ch 04: Intro to Forces (Dynamics)||3hrs & 42mins||0% complete||WorksheetStart|
|Ch 05: Friction, Inclines, Systems||4hrs & 32mins||0% complete||WorksheetStart|
|Ch 06: Centripetal Forces & Gravitation||3hrs & 39mins||0% complete||WorksheetStart|
|Ch 07: Work & Energy||3hrs & 55mins||0% complete||WorksheetStart|
|Ch 08: Conservation of Energy||6hrs & 54mins||0% complete||WorksheetStart|
|Ch 09: Momentum & Impulse||5hrs & 35mins||0% complete||WorksheetStart|
|Ch 10: Rotational Kinematics||3hrs & 4mins||0% complete||WorksheetStart|
|Ch 11: Rotational Inertia & Energy||7hrs & 7mins||0% complete||WorksheetStart|
|Ch 12: Torque & Rotational Dynamics||2hrs & 9mins||0% complete||WorksheetStart|
|Ch 13: Rotational Equilibrium||4hrs & 10mins||0% complete||WorksheetStart|
|Ch 14: Angular Momentum||3hrs & 6mins||0% complete||WorksheetStart|
|Ch 15: Periodic Motion (NEW)||2hrs & 17mins||0% complete||WorksheetStart|
|Ch 15: Periodic Motion (Oscillations)||3hrs & 16mins||0% complete||WorksheetStart|
|Ch 16: Waves & Sound||3hrs & 25mins||0% complete||WorksheetStart|
|Ch 17: Fluid Mechanics||4hrs & 39mins||0% complete||WorksheetStart|
|Ch 18: Heat and Temperature||4hrs & 9mins||0% complete||WorksheetStart|
|Ch 19: Kinetic Theory of Ideal Gasses||1hr & 40mins||0% complete||WorksheetStart|
|Ch 20: The First Law of Thermodynamics||1hr & 49mins||0% complete||WorksheetStart|
|Ch 21: The Second Law of Thermodynamics||4hrs & 56mins||0% complete||WorksheetStart|
|Ch 22: Electric Force & Field; Gauss' Law||3hrs & 35mins||0% complete||WorksheetStart|
|Ch 23: Electric Potential||1hr & 45mins||0% complete||WorksheetStart|
|Ch 24: Capacitors & Dielectrics||1hr & 48mins||0% complete||WorksheetStart|
|Ch 25: Resistors & DC Circuits||3hrs & 22mins||0% complete||WorksheetStart|
|Ch 26: Magnetic Fields and Forces||2hrs & 25mins||0% complete||WorksheetStart|
|Ch 27: Sources of Magnetic Field||2hrs & 30mins||0% complete||WorksheetStart|
|Ch 28: Induction and Inductance||2hrs & 48mins||0% complete||WorksheetStart|
|Ch 29: Alternating Current||2hrs & 37mins||0% complete||WorksheetStart|
|Ch 30: Electromagnetic Waves||1hr & 12mins||0% complete||WorksheetStart|
|Ch 31: Geometric Optics||3hrs||0% complete||WorksheetStart|
|Ch 32: Wave Optics||1hr & 15mins||0% complete||WorksheetStart|
|Ch 34: Special Relativity||2hrs & 10mins||0% complete||WorksheetStart|
|Ch 35: Particle-Wave Duality||Not available yet|
|Ch 36: Atomic Structure||Not available yet|
|Ch 37: Nuclear Physics||Not available yet|
|Ch 38: Quantum Mechanics||Not available yet|
|What is a Wave?||27 mins||0 completed|
|The Mathematical Description of a Wave||36 mins||0 completed|
|Waves on a String||24 mins||0 completed|
|Wave Interference||17 mins||0 completed|
|Standing Waves||24 mins||0 completed|
|Sound Waves||13 mins||0 completed|
|Standing Sound Waves||13 mins||0 completed|
|Sound Intensity||25 mins||0 completed|
|The Doppler Effect||17 mins||0 completed|
|Beats||10 mins||0 completed|
Concept #1: Waves On A String
The correct equation should be f = v/λ, not what I wrote.
The correct answer is f = v/λ = 4.9 / 0.12 = 40.8 Hz
Hey guys, in this video we're going to talk about waves on strings. If you were to fix a wave at one end against something like a wall, grab the other and then just whip it up and down you would see that you'd produce a bunch of waves travelling down the length of that string. So this is what we want to talk about now. As we've said before a wave on a string is a very common example of a transverse wave. One whose oscillations are perpendicular to its propagation and we explained why this was transverse before but if you whip the string up and down you're going to produce these wave pulses that travel towards the wall. These waves that travel towards the wall. Now a wave on a string is no different really than any other wave in a lot of respects. It has the same mathematical description, it has the same characteristics, an amplitude, a propagation speed, a wavelength, frequency, etc. but there are a couple things that are unique to each wave. The specific energy carried by that wave is going to be unique to the wave and the specific propagation speed of that wave is going to be unique to the wave. Remember that propagation speed depends upon two things, the type of wave and the characteristics of the medium that the wave is in. So what we're going to focus ourselves on in this video is the speed of this wave and of course we're talking about the propagation speed. Alright the propagation speed of a wave on a string is given by the following equation. The square root of T over mu where T is the tension in the string and mu is the mass per unit length of that string.
Let's do a quick example. A 1.2 meter string has a tension of 100 Newtons. If it has a mass of 5 kilogram's, what is the frequency of a wave with a 12 centimetre wavelength on this string? Well we know we want to find the frequency of the wave and we are told the wavelength. Since we are dealing with both frequency and wavelength we're probably going to be using the equation V equals lambda F and we can easily solve for F by dividing lambda over. So the only question is what's V? The speed on the string. Well we know that for waves on a string the speed is just the tension divided by the mass per unit length. We know the tension, we know the length and we know the mass so we know everything we need to solve this problem. This is going to be 100 Newtons of tension divided by the mass per unit length. So it's 5 kilogram's in mass and it's 1.2 meters in length. So that whole thing is going to be about 4.9 meters per second. This means that the frequency is 0.12 meters, it was given in centimetres but I need to convert to meters, divided by 4.9 meters per second which equals 0.024 Hertz. Alright, no big deal. That wraps up our discussion on our introduction to waves on a string. Thanks for watching guys.
Example #1: Unknown Mass Of A String
Hey guys, let's do a quick example. An easy way to find the mass of a string is to produce a wave on it. A 5 centimeter string of unknown mass produces a wave with a 12 centimeter wavelength that has a 70 Hertz frequency when its tension is 10 Newtons. Given this, what must the mass of the spring be? Sorry of the string be. Well the only equation we have so far that deals with the mass on a string is the speed equation. Speed of waves on a string are given by the square root of the tension divided by the mass per unit length. This is the tension divided by the mass per unit length.
What we want is to isolate that mass. So we can just rewrite this as the square root, sorry, the squared of the speed is T over M over L. Take M over L and multiply it into the numerator, take V squared and divide it into the denominator. This M over L is T over V squared and then lastly take L multiply it into the numerator and we have M is TL over V squared. So what we need to know is V squared. Well the relationship for speed of any wave is that it's equal to the wavelength times the frequency. We know that it's a 12 centimeter wavelength and a 70 Hertz frequency. So this is going to be 0.12, we want it in meters, times 70 which is 8.4 meters per second. Now we can solve for the mass. So the mass is going to be TL over V squared, the tension is 10 Newtons, the length is 5 centimeters or 0.05 meters and the speed we found was 8.4 and we need that squared and that comes out to 0.007 kilograms or if I reduce that by a factor of 1000 that's 7 grams. Either of these answers are perfectly fine. Alright guys, that wraps up this problem. Thanks for watching
Practice: A 2 m long string is arranged as shown in the figure below. If the mass of the string is 0.5 kg, what would the frequency be of a wave with a 2 cm wavelength?
Example #2: Speed Of A Wave Traveling Up A String
Hey guys, let's do a quick example. A string is hung vertically from an anchor point. If you were to grasp the string at the bottom and shake it to produce a wave traveling up the string, what happens to the speed of the wave as it moves up the string? So we have some anchor point here, a rope hanging, we grab that rope and we produce a wave on it by whipping it. So this is travelling upwards. What happens to the speed of that wave as it travels up the rope? A lot of people are probably going to want to say it decreases right off the bat but don't associate this with freefall motion. We know the speed of a wave on the string equals the tension divided by the mass per unit length. The mass per unit length is not changing but the tension in the rope is. Tension in a rope is always increasing the higher up the rope you get. The simple reason is this, imagine the rope was built up, was broken up into a bunch of chains, chain links.
How much tension is in the chain link second from the bottom? Well all that's holding, all that it's supporting is one little MG which I'll say is the weight of a single link so the tension is just one little MG but what if we go 1, 2, 3, 4, 5 up? Then there's four chains below it so what's the weight it's supporting? 4 little MG so its tension is 4 little MG. Let's go up to the top of this chain link. It has 1, 2, 3, 4, 5, 6, 7, 9 so the weight that it's is 9 little MG so the tension is 9 little MG so clearly the tension in the rope increases the higher you go up because there's more rope being supported the higher you go up. So as the tension goes up, the speed goes up. So the speed of the wave actually increases as it goes up the rope. This is contrary to our initial sort of visceral guess that it should actually decrease on its way up like objects in freefall. Alright guys, that wraps up this question. Thanks for watching.
Concept #2: Energy Carried By Waves On A String
calculator error --> answer is 140 Hz
Hey guys, in this video we want to talk about the energy carried by waves on a string. Alright let's get to it. Waves on a string carry both kinetic and potential energy. They obviously carry kinetic because as a wave which has, sorry, as a wave on a string that string has mass as it propagates it has a velocity, it has mass so it has kinetic energy but when the string is just flat, static, no waves on it when you whip it that whip stretches the string a little bit to produce a wave. That wave has some stretchiness in that string so it also carries some potential energy with it as well. The kinetic energy is due to the motion of the string, sorry wrong color, kinetic energy is due to the motion of the string and the potential energy is due to the stretching of the string. This energy depends upon three things. It depends upon the frequency of the wave, it depends upon the amplitude of the wave and it depends upon the wavelength of the wave so pretty much the three main characteristics of a wave. Now it turns out that the kinetic energy and the potential energy are the same for a wave on a string. They are both given by the same equation given right here in our orange box where the variables mu, omega, A and lambda are the same as they have been so far. The total energy per unit wavelength is just going to be the sum of the kinetic energy and the potential energy per unit wavelength. What I mean by per unit wavelength is this is the energy carried by a single wavelength. The energy per unit wavelength is just double this, it's this plus itself so it's one half mu omega squared, A squared, lambda.
Alright, let's do a quick example. A string 5 meters in length carries a wave along its length. If the tension is 50 Newtons and the mass per unit length is 0.005 kilograms per meter, how much kinetic energy does the string carry for a wave with a frequency of 50 Hertz and an amplitude of 1 centimeter? So we want to find first the energy per wavelength then we'll find the total energy. This is one half mu omega squared A squared lambda. So we know mu, mu was given to us the mass per unit length. The frequency is given to us so we can find the angular frequency pretty easily. The amplitude is given to us, the only thing that's not given to us is the wavelength. The wavelength we do have to calculate using our wave equation. This means that the wavelength is F over V. The question is what's the speed of the wave? Well since this is a wave on a string the speed is just the tension divided by the mass per unit length square rooted. The tension as we're told is 50 Newtons, the mass per unit length is 0.005 and this whole thing equals 10, sorry, 100 meters per second. The answer under the square root is 10000, the square root of 10000 is 100. So this means that the wavelength is 50 Hertz, the frequency divided by 100 meters per second which is, sorry this is backwards I messed up on this equation right here. This equation is V divided by F not F divided by V. So this is 100 divided by 50 which is 2 meters. Now we have everything we need to know to solve for the energy per unit wavelength. That's one half 0.005 was the mass per unit length, the angular frequency is just 2 pi times 50 Hertz was the frequency squared times the amplitude 0.01 meters squared, everything needs to be in SI units, times the wavelength which was 2 meters and this is 0.49 joules but this is not what the question asked. The question asked how much energy is carried by the string? Now this is the energy for a single wavelength, so the question is how many wavelengths does the string have? Well the string is 5 meters in length and if the wavelength is 2 meters then the whole string carries 2 and a half wavelengths so the total energy is going to be 2.5, sorry, 2 and a half times the energy per unit wavelength which is 1.22 joules. Now something else you could ask is how much power is carried by the string? Well we know the energy per unit wavelength and how long does it take for a wavelength to pass? Just a period. So if I divide the energy per unit wavelength by the period I get the power and what we have is we have this last lambda divided by the period which gives us the velocity and this is the power period this isn't the power per unit wave length this is the total power.
If you for instance asked for 2 times the energy or the energy per 2 wavelengths you have to divide it by 2 periods so that 2 cancels from the numerator and the denominator. This power is not dependent on number of wavelength this is just the power. So a 0.7 meter string supports a wave of 0.05 meter amplitude. If the mass of the string is 0.05 kilograms and the tension in the string is 15 Newtons what must the frequency of the wave be to carry 100 watts of power? Since we're dealing with power we'll start there. This is one half mu omega squared A squared times V. We know the tension in the string, we know the mass per unit we know the mass of the string and we know the length of the string so we can easily find the speed. The speed is just the square root of T divided by mu which is the square root of 15 Newtons divided by 0.05 kilograms divided by 0.7 that's the mass per unit length and that is 14.5 meters per second. Now what we want to know is the frequency, what must the frequency be to have 100 watts of power? So we need to rearrange our power equation to solve frequency. The frequency in this equation is actually the angular frequency but we can always convert that to linear frequency later so this becomes omega squared is 2P, I multiply the 2 up the 2 from the denominator up and everything else goes into the denominator, mu A squared V. So omega is the square root of 2P over mu A squared V which is going to be 2 times the power's 100 watts, the mass per unit length as I said was the mass, 0.05 kilograms, divided by the length, 0.7 meters, the amplitude we're told is 0.05 meters and the speed we found as 14 and a half. So this whole thing becomes 879 inverse seconds. The last thing we need to do is we need to take this number which is angular frequency and convert it into linear frequency. Don't forget that linear frequency is angular frequency divided by 2 pi, so this is 879 divided by 2 pi which is about 280 Hertz. Alright, that wraps up our discussion into the energy carried by waves on a string. Thanks for watching guys.
Example #3: Maximum Frequency Of A Wave On A String
Hey guys, I hope you're able to solve this problem on your own. If not here's a little bit of help. Waves are produced on a string by a paddle that can move up and down shaking the free end of a string to produce a 5 centimeter amplitude wave as shown in the figure below. The string has a mass per unit length of half a kilogram per meter and a tension of 100 Newtons. If the maximum power the paddle can deliver is 50 watts, what is the greatest frequency, sorry, what is the greatest frequency wave the string can support? So remember that power carried by a wave on a string is one half times mu omega squared A squared V.
So if we want to know frequency we'll solve for omega squared multiply the 2 up to the numerator, everything else goes to the denominator and then take the square root of both sides. We're told the maximum power, we're told the mass per unit length, we're told the amplitude, all we're missing is the speed of this wave. Now this is a wave on a string so the speed is just going to be the square root of the tension divided by the mass per unit length. We're told that the tension's 100 Newtons so this is 100 divided by half a kilogram per meter and that speed is 14 meters per second. Now we know everything we need to solve for that angular frequency. This is 2 times the power which we said was 50 watts, divided by the mass per unit length which is half a kilogram per meter, the amplitude was 5 centimeters or 0.05 meters squared and the speed which was 14 meters per second and this whole thing is going to be 76 inverse seconds but we're not done remember we're looking for the greatest frequency that means linear frequency so we need one more step which says that the linear frequency is omega divided by 2 pi which is 76 inverse seconds over 2 pi which is going to be 12 Hertz. Alright and that wraps up this problem. Thanks for watching guys
Enter your friends' email addresses to invite them: