|Ch 01: Units & Vectors||2hrs & 22mins||0% complete||WorksheetStart|
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|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|
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|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|
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|Ch 19: Kinetic Theory of Ideal Gasses||1hr & 40mins||0% complete||WorksheetStart|
|Ch 20: The First Law of Thermodynamics||1hr & 49mins||0% complete||WorksheetStart|
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|Ch 22: Electric Force & Field; Gauss' Law||3hrs & 32mins||0% complete||WorksheetStart|
|Ch 23: Electric Potential||1hr & 52mins||0% complete||WorksheetStart|
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|Ch 26: Magnetic Fields and Forces||2hrs & 25mins||0% complete||WorksheetStart|
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|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|
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|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: Standing Waves
Hey guys in this video we're going to talk about standing waves which are types of waves that occur due to interference between overlapping waves. Alright let's get to it. Remember that waves along a string that are fixed at a boundary are going to be reflected backwards as inverted waves. So these waves right here were a bunch of waves that had been produced by this hand sent forward and have already reflected there inverted and remember that these inverted waves will have the same frequency as the incident waves as the initial waves that hit the boundary. Now imagine holding the end of a string fixed to a wall and vibrating it. We're going to go up and down we're just going to shake it, we're going to produce a bunch of pulses that are going to travel from the free end towards the fixed end and they're going to reflect backwards these inverted reflections are going to have the same frequency as they reflect off and then they're going to encounter new pulses that are being produced. Forward pulses and the reflections since they're going to be in the same space at the same time are going to interfere. Now normally what's going to be produced is some sort of uncoordinate vibration, essentially nothing just a string that's doing a whole bunch of this but it's specific frequencies you are going to produce what are known as standing waves. I drew one instance of a standing wave here and another instance of a standing wave here. The first standing wave that I drew is a wave that essentially looks like this and if you go faster one part is going to go down as the other is going to go up like this that's the second one that I drew these are called standing waves because they appear to be standing still instead of individual wave pulses that are moving forward we now have just a pulse that's going up and down or going up and down alternating like this I don't have three hands so I can really do another standing wave which would be formed of three pulses. Two types of standing waves exist node node standing waves and node antinode standing waves in different scenarios produce either node node or node antinode what a node is is it's a point of no displacement, I highlight I bolded and underscored this right there so that'll help you remember what a node is node no displacement. An antinde is the opposite of a node it's just a point of maximum displacement so if you look at the standing waves drawn this is a node and this is a node this is what we would call a node node standing wave because both ends of the string are no nodes and then this point is an antinode, here we have 3 nodes a node a node a node another node this is still a node node standing wave because both ends of the string are at nodes and we have 2 anitnodes. Now let's look at the differences between node node and node antinode standing waves. For node node standing waves as I said both ends are nodes for node anitnode standing waves one end is a node the other is an antinode, so I drew pictures of the three largest wavelength or lowest frequency of each if you see here on either end the wave returns back to the horizontal axis returns back to a node here on one end is a node but on the other end it's always going to be at the amplitude you can see here this is the amplitude it's the point of maximum displacement because of that wavelengths and frequencies have to abide by equations for a node node standing waves the allowed wavelengths are given by this equation where N is any integer and the allowed frequencies are given by this equation where N is any integer as long as your wavelength or your frequency agrees with those equations it's a standing wave if you have a frequency for instance if you wipe your hand at a frequency that does not follow this equation you will not produce a standing wave you're just going to produce a bunch of uncoordinated vibrations that don't mean anything I am going to minimise myself for node antinode standing waves the allowed wavelengths obey this equation but N has to be odd this is crazy important to remember that N which is called the harmonic number we call N the harmonic number is any integer for node node standing waves but it has to be odd you can only have odd harmonic numbers for node antinode standing waves and the allowed frequencies for node antidone standing waves are given by this equation where once again the harmonic number has to be odd that's crazy important that you remember that the harmonic number must be odd.
Alright let's do a quick example to finish this off. You want to produce standing waves on the string shown in the figure below by vibrating the end you're grasping what is the third largest frequency producible on the string if it has a mass of 0.05 kilogram's and a tension of 150 newtons assume there's no friction between the ring at the end of the string and the pole what kind of standing waves do you think these are going to be guys your hand is going to be a node but this end which is free to move is going to be an antinode. So this is a node antinode if we want the third largest frequency we have to use this equation. What's the third largest harmonic number for a node antinode? Remember N has to be odd so it's 1,3,5,7 etc. The third largest is 5 it's not 3 because 2 doesn't count the third largest is 5 so the fifth harmonic which is the third largest harmonic in this case is 5 V over 4 L and all we have to do is figure out what V is, the speed of this wave on the string we know that the string has a mass of 0.05 kilogram's and a tension of 150 newtons and we know its length is 15 centimetres so the speed is going to be the square root of T over mu, which is the square root of 150 newtons the tension divided by 0.05 the mass over the length 0.15. Right and that's going to be 21.2 meters per second now that we know the speed we can solve for the frequency. This is just going to be , 21.2 over 4 the length is 0.15 and so that is 177 hertz.Alright guys that wraps up this introduction to standing waves. Thanks for watching.
Practice: In the following figure, what is the harmonic number of the standing wave? The wavelength of the standing wave? If the frequency of the standing wave is 30 Hz, what is the speed of the waves producing the standing wave?
Example #1: Unknown Harmonic Frequency
Hey guys let's do an example a string fixed at both ends produces a harmonic frequency of 650 Hertz with the next highest harmonic frequency being 780 Hertz if the same string can produce a harmonic frequency of 390 Hertz, what harmonic frequency is the next highest? So we have two different pairs of harmonic frequencies initially we're saying some harmonic frequency which I will call F. N. which is N V over 2 L right it's fixed at both ends. So this is node node so I'm using N.V over 2 L which is our node node equation. This is 650 Hertz, now I'm saying the next highest which is N plus 1 right if N was 650 then the next highest is N plus 1 that's going to be N plus 1 V over 2 L, which is 780 hertz. But notice something I can break up that N plus 1 into 2 separate fractions I can say that N V over 2 L plus V over 2 L equals 780 hertz what's key here is that this term right here V over 2 L that is how much the frequency grows every time you go up a new harmonic number so if I'm at 17 and I go up to 18 it's going to grow by V over 2 L so V over 2 L is exactly what I need to know to solve the problem and luckily I know what N V over 2 L is right. It's just 650 hertz, so this tells me 650 hertz plus V over 2 L which is our step in the frequency it's how much the frequency increases by when we increase our harmonic number by 1. That's 780 hertz, so V over 2 L is 780 minus 650 which is a 130 hertz. So this is sort of the linchpin to solving that problem you need to know the step we don't know anything about the length of the string the tension the mass in the string the speed we don't know any of that so we can't actually solve for V over 2 L but we found it by comparing 1 arbitrary harmonic frequency to the next highest now we're at some other harmonic frequency some other unknown it really doesn't matter I'll call it M and we're saying F of M. which is M V over 2 L is 390 hertz.
Now the next one above that is M plus 1, so this is F M plus 1 which is M plus 1 V over 2 L and once again we can split up this fraction M V over 2 L plus V over 2 L. So you see there we've exposed this is that increase in frequency whenever you increase the harmonic number by 1 this was our original harmonic frequency at M. Which was 390 hertz so this become 390 hertz plus that step frequency that we found plus 130 hertz. Which is 520 hertz and you see we didn't know anything about this problem we didn't know how long the string was we didn't know the speed of the string we didn't even know what these harmonic numbers were but all of that is irrelevant because there was enough information to solve for that step that increase in harmonic frequency every time you increase the harmonic number by 1. Alright guys thanks for watching.
Example #2: Standing Wave On A Guitar
Hey guys, let's do an example. A guitar is played by plucking the string which emits a sound at the frequency it vibrates. To play different tunes, the string is held down by your finger at what are called fret locations. Each subsequent fret shortening the length of the string by about 2 centimeters. If a guitar string has a full length of 64 centimeters and a tension of 70 Newtons and a mass per unit length of 1 times 10 to the -3 kilograms per meter, what is the frequency emitted by the guitar if held at the 10th fret location?
So remember the string starts out at 64 centimeters and each fret shortens the length of the string by about 2 centimeters. So if you're at the 10th fret, that means that you're at -20 centimeters for the length. So the new effective length which I'll call L prime is 64 centimeters minus 20 centimeters which is 44 centimeters. So you're actually playing a string that's only 44 centimeters long because you're holding one end of the string down to shorten the total length of the string. Now the string is fixed at both ends so this is a node-node standing wave which means that the frequency is going to be N times V over 2 times our new length not the 64 centimeters because when you're holding it at the 10th fret you've effectively shortened the length of the guitar. The question is what mode are we in? What harmonic number?
Well if the guitar string is fixed at both ends and you pluck it what happens is it vibrates like this, up and down. This is N equals one and that's all we need. So the frequency is going to be V over 2 times the new length. That speed we don't quite know but we know enough to solve for it. The speed is the square root of the tension which is 70 divided by the mass per unit length which is 1 times 10 to the -3 which is 26, sorry, 265 meters per second. Couldn't read my notes for a second. 265 meters per second. So this is going to be 265 meters per second divided by 2 times our new length which is 44 centimeters or 0.44 which is going to be about 301 Hertz. That's going to be about the frequency emitted by the guitar. Alright guys, that wraps up this problem. Thanks for watching.
Practice: An unknown mass hangs on the end of a 2 m rope anchored to the ceiling when a strong wind causes the rope to vibrate and hum at its fundamental frequency of 100 Hz. If the rope has a mass of 0.15 kg, what is the unknown mass?
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