1 00:00:00,336 --> 00:00:01,351 - [Lecturer] Hooke's Law tells you how to find 2 00:00:01,351 --> 00:00:04,976 the force exerted by an ideal or linear spring. 3 00:00:04,976 --> 00:00:05,916 And it's a simple law. 4 00:00:05,916 --> 00:00:07,699 It tells you the amount of force that spring 5 00:00:07,699 --> 00:00:09,949 is gonna exert will be proportional to the amount 6 00:00:09,949 --> 00:00:12,701 that spring has been stretched or compressed 7 00:00:12,701 --> 00:00:15,341 from its equilibrium or natural length. 8 00:00:15,341 --> 00:00:17,723 Which, in equation form just says that the magnitude 9 00:00:17,723 --> 00:00:20,509 of the spring force, is gonna equal the spring constant 10 00:00:20,509 --> 00:00:22,990 multiplied by the amount the spring has been stretched 11 00:00:22,990 --> 00:00:24,067 or compressed. 12 00:00:24,067 --> 00:00:26,988 Note this x is not the length of the spring. 13 00:00:26,988 --> 00:00:29,743 The x is how much that spring has been stretched from 14 00:00:29,743 --> 00:00:32,849 or compressed from the equilibrium position 15 00:00:32,849 --> 00:00:34,889 or the unstretched position. 16 00:00:34,889 --> 00:00:36,053 So what's an example problem 17 00:00:36,053 --> 00:00:37,991 involving Hooke's law look like? 18 00:00:37,991 --> 00:00:39,363 Let's say an ideal spring is hanging 19 00:00:39,363 --> 00:00:41,046 from the ceiling at rest, 20 00:00:41,046 --> 00:00:43,520 and it has an unstretched length L1. 21 00:00:43,520 --> 00:00:46,322 And then you hang a mass M from the spring at rest 22 00:00:46,322 --> 00:00:49,108 and it stretches the spring to a length L2. 23 00:00:49,108 --> 00:00:52,201 What is an expression for the spring constant of the spring? 24 00:00:52,201 --> 00:00:53,218 So the force of gravity has to 25 00:00:53,218 --> 00:00:55,291 be balanced by the spring force. 26 00:00:55,291 --> 00:00:57,043 That means the magnitude of the spring force 27 00:00:57,043 --> 00:00:59,438 is equal to the magnitude of the force of gravity. 28 00:00:59,438 --> 00:01:01,766 The spring force is always K times X. 29 00:01:01,766 --> 00:01:03,452 What is X gonna represent? 30 00:01:03,452 --> 00:01:05,924 It's not gonna be L1 or L2. 31 00:01:05,924 --> 00:01:08,229 X is how much the spring has been stretched 32 00:01:08,229 --> 00:01:12,744 from its equilibrium position which would be L2 minus L1. 33 00:01:12,744 --> 00:01:17,133 And if we solve for K we get mg over L2 minus L1. 34 00:01:17,133 --> 00:01:19,577 What's a simple harmonic oscillator? 35 00:01:19,577 --> 00:01:22,489 A simple harmonic oscillator's any variable who's change 36 00:01:22,489 --> 00:01:25,698 can be described by a sine or cosine function. 37 00:01:25,698 --> 00:01:27,031 What does that function look like? 38 00:01:27,031 --> 00:01:28,097 It looks like this. 39 00:01:28,097 --> 00:01:30,687 So the variable that's changing is a function of time, 40 00:01:30,687 --> 00:01:33,933 which could be the vertical position of a mass on a spring. 41 00:01:33,933 --> 00:01:36,289 The angle of a pendulum or any other 42 00:01:36,289 --> 00:01:38,637 simple harmonic oscillator is gonna be equal to 43 00:01:38,637 --> 00:01:40,622 the amplitude of the motion, 44 00:01:40,622 --> 00:01:43,701 which is the maximum displacement from equilibrium, 45 00:01:43,701 --> 00:01:47,344 times either sine or cosine of two pie 46 00:01:47,344 --> 00:01:49,037 times the frequency of the motion 47 00:01:49,037 --> 00:01:50,813 times the variable t. 48 00:01:50,813 --> 00:01:52,764 Which, since frequency is one over the period, 49 00:01:52,764 --> 00:01:55,153 you could write as two pie over the period 50 00:01:55,153 --> 00:01:57,187 times the time variable t. 51 00:01:57,187 --> 00:01:59,416 How do you know whether you use sine or cosine? 52 00:01:59,416 --> 00:02:01,810 Well sine starts at zero and goes up. 53 00:02:01,810 --> 00:02:04,764 And cosine starts at a maximum and goes down. 54 00:02:04,764 --> 00:02:05,740 So if you know the behavior 55 00:02:05,740 --> 00:02:08,047 of your oscillator at t equals zero. 56 00:02:08,047 --> 00:02:11,293 You can decide whether to use cosine or sine. 57 00:02:11,293 --> 00:02:13,068 Something that's important to know is how to find 58 00:02:13,068 --> 00:02:14,642 the period of an oscillator. 59 00:02:14,642 --> 00:02:16,608 The period of a mass on a spring is gonna be 60 00:02:16,608 --> 00:02:19,615 two pie times the square root, the mass connected 61 00:02:19,615 --> 00:02:22,996 to the spring divided by the spring constant k. 62 00:02:22,996 --> 00:02:25,943 Note that this does not depend on the amplitude. 63 00:02:25,943 --> 00:02:29,022 If you stretch that mass farther, it'll go faster 64 00:02:29,022 --> 00:02:31,137 and it has farther to go which cancels out 65 00:02:31,137 --> 00:02:33,075 and the period remains the same. 66 00:02:33,075 --> 00:02:35,072 And the formula for the period of a pendulum, 67 00:02:35,072 --> 00:02:37,081 which is a mass swinging on a string, 68 00:02:37,081 --> 00:02:38,946 is gonna be two pie times the square root 69 00:02:38,946 --> 00:02:41,888 the length of the string, divided by the magnitude 70 00:02:41,888 --> 00:02:43,817 of the acceleration due to gravity. 71 00:02:43,817 --> 00:02:46,291 Which also does not depend on the amplitude, 72 00:02:46,291 --> 00:02:48,568 as long as the angles are small. 73 00:02:48,568 --> 00:02:51,062 And it doesn't depend on the mass either. 74 00:02:51,062 --> 00:02:53,061 How do you find this period on a graph? 75 00:02:53,061 --> 00:02:54,768 Well if you're given the graph of the motion 76 00:02:54,768 --> 00:02:57,266 of a simple harmonic oscillator as a function of time, 77 00:02:57,266 --> 00:03:01,216 the interval between peaks is gonna represent the period. 78 00:03:01,216 --> 00:03:04,478 Or the time is takes for this oscillator to reset. 79 00:03:04,478 --> 00:03:05,858 So what's an example problem involving 80 00:03:05,858 --> 00:03:07,742 simple harmonic motion look like? 81 00:03:07,742 --> 00:03:10,014 Let's say in a lab a mass M on Earth 82 00:03:10,014 --> 00:03:12,250 can either be hung on a string of length L 83 00:03:12,250 --> 00:03:13,827 and allowed to swing back and forward 84 00:03:13,827 --> 00:03:17,232 with a period T pendulum, or hung on a spring of spring 85 00:03:17,232 --> 00:03:19,539 constant k and allowed to oscillate up and down 86 00:03:19,539 --> 00:03:21,280 with a period T spring. 87 00:03:21,280 --> 00:03:24,985 If a 2M mass were used instead of the 1M mass, 88 00:03:24,985 --> 00:03:27,383 what would happen to the period of the two motions? 89 00:03:27,383 --> 00:03:30,650 Well the period of a pendulum doesn't depend on the mass. 90 00:03:30,650 --> 00:03:32,974 And so the period of the pendulum would not change. 91 00:03:32,974 --> 00:03:34,552 The answer would have to be D. 92 00:03:34,552 --> 00:03:36,074 What are waves? 93 00:03:36,074 --> 00:03:38,975 Waves are disturbances that travel through a medium 94 00:03:38,975 --> 00:03:43,155 and transfer energy in momentum over significant distances 95 00:03:43,155 --> 00:03:47,367 without transmitting any mass itself over those distances. 96 00:03:47,367 --> 00:03:48,628 What does medium mean? 97 00:03:48,628 --> 00:03:50,977 This is a fancy word for the material through which 98 00:03:50,977 --> 00:03:52,530 the wave can travel. 99 00:03:52,530 --> 00:03:54,971 So you can classify a wave by the medium it's in. 100 00:03:54,971 --> 00:03:57,208 But you could also classify waves by the type 101 00:03:57,208 --> 00:03:59,096 of disturbance you've created. 102 00:03:59,096 --> 00:04:01,459 For transverse waves, the disturbance of the medium 103 00:04:01,459 --> 00:04:04,580 is perpendicular to the wave velocity. 104 00:04:04,580 --> 00:04:07,130 By wave velocity, we mean the direction in which 105 00:04:07,130 --> 00:04:09,129 the disturbance travels. 106 00:04:09,129 --> 00:04:10,694 And by oscillation of the medium, 107 00:04:10,694 --> 00:04:13,483 we mean the direction in which the particles of the medium 108 00:04:13,483 --> 00:04:14,860 actually move. 109 00:04:14,860 --> 00:04:17,541 For a wave on a string, the particles move up and down 110 00:04:17,541 --> 00:04:19,889 but the disturbance travels to the right. 111 00:04:19,889 --> 00:04:21,702 So this would be a transverse wave. 112 00:04:21,702 --> 00:04:24,558 For longitudinal waves, the oscillation or disturbance 113 00:04:24,558 --> 00:04:27,505 of the medium is parallel to the wave velocity. 114 00:04:27,505 --> 00:04:29,627 The classic longitudinal wave is sound. 115 00:04:29,627 --> 00:04:32,263 If a sound wave were traveling rightward through the air, 116 00:04:32,263 --> 00:04:34,228 it would look like a compressed region, 117 00:04:34,228 --> 00:04:36,749 and the air itself would move back and forth, 118 00:04:36,749 --> 00:04:39,229 right and left, parallel to the direction 119 00:04:39,229 --> 00:04:41,289 the wave disturbance is traveling. 120 00:04:41,289 --> 00:04:43,736 Which makes sound waves longitudinal. 121 00:04:43,736 --> 00:04:47,414 For every type of wave, the speed of that wave disturbance 122 00:04:47,414 --> 00:04:50,082 is gonna be equal to the wavelength of the wave 123 00:04:50,082 --> 00:04:51,531 divided by the period. 124 00:04:51,531 --> 00:04:53,730 In other words, if you watched a wave crest, 125 00:04:53,730 --> 00:04:57,271 that wave crest would move one wavelength every period. 126 00:04:57,271 --> 00:04:59,577 And since the speed is distance per time, 127 00:04:59,577 --> 00:05:02,216 the speed of the wave crest would be one wavelength 128 00:05:02,216 --> 00:05:03,253 per period. 129 00:05:03,253 --> 00:05:06,445 You could find the wavelength on a graph of y versus x 130 00:05:06,445 --> 00:05:08,889 by finding the distance between crests. 131 00:05:08,889 --> 00:05:10,903 And if you're wondering why this doesn't represent 132 00:05:10,903 --> 00:05:11,746 the period? 133 00:05:11,746 --> 00:05:14,561 It's because this is a graph of the wave versus x, 134 00:05:14,561 --> 00:05:17,499 versus the horizontal position, not the time. 135 00:05:17,499 --> 00:05:20,426 You could make a graph of y versus the time. 136 00:05:20,426 --> 00:05:22,202 And what that would represent is the motion 137 00:05:22,202 --> 00:05:26,010 of a single point on the wave for all moments in time. 138 00:05:26,010 --> 00:05:28,240 And for this graph versus time, 139 00:05:28,240 --> 00:05:30,699 the interval between peaks is the period. 140 00:05:30,699 --> 00:05:32,259 So if you get a graph of a wave, 141 00:05:32,259 --> 00:05:35,519 you've gotta check whether it's versus x or versus t. 142 00:05:35,519 --> 00:05:38,153 If it's versus x, peak to peak is the wavelength. 143 00:05:38,153 --> 00:05:41,057 And if it's versus t, peak to peak is the period. 144 00:05:41,057 --> 00:05:43,829 And since one over the period is equal to the frequency, 145 00:05:43,829 --> 00:05:45,550 we can rewrite this speed formula 146 00:05:45,550 --> 00:05:49,054 as the speed of a wave equals the wavelength of a wave 147 00:05:49,054 --> 00:05:50,349 times the frequency. 148 00:05:50,349 --> 00:05:51,882 And the way it's given on the formula sheet 149 00:05:51,882 --> 00:05:54,397 on the AP exam is that the wavelength of a wave 150 00:05:54,397 --> 00:05:57,586 is equal to the speed of the wave divided by the frequency. 151 00:05:57,586 --> 00:05:59,956 Now this formula confuses a lot of people though, 152 00:05:59,956 --> 00:06:03,056 because they think if you increase the frequency, 153 00:06:03,056 --> 00:06:05,053 that'll increase the speed of the wave. 154 00:06:05,053 --> 00:06:06,086 But that's not true. 155 00:06:06,086 --> 00:06:08,162 Increasing the frequency will cause the wavelength 156 00:06:08,162 --> 00:06:11,428 to decrease and the speed of the wave will remain constant. 157 00:06:11,428 --> 00:06:13,916 The only way to change the speed of a wave 158 00:06:13,916 --> 00:06:16,882 is to change the properties of the medium itself. 159 00:06:16,882 --> 00:06:18,786 In other words the only way to change the speed 160 00:06:18,786 --> 00:06:21,051 of the waves on water would be to change something 161 00:06:21,051 --> 00:06:24,841 about the water itself, its density, its salinity, 162 00:06:24,841 --> 00:06:26,672 the temperature of that water. 163 00:06:26,672 --> 00:06:29,619 Changing frequency isn't gonna change the speed of the wave. 164 00:06:29,619 --> 00:06:32,096 And neither will changing the amplitude. 165 00:06:32,096 --> 00:06:34,063 The only thing that changes the speed of the wave 166 00:06:34,063 --> 00:06:36,620 is changes to the medium itself. 167 00:06:36,620 --> 00:06:39,079 And so what's an example problem involving waves look like? 168 00:06:39,079 --> 00:06:41,879 Let's say a sound lab is being conducted in a lab room 169 00:06:41,879 --> 00:06:44,763 with total cubic volume V and temperature T. 170 00:06:44,763 --> 00:06:47,206 A speaker in the room is hooked up to a function generator 171 00:06:47,206 --> 00:06:50,564 and it plays a note with frequency f and amplitude A. 172 00:06:50,564 --> 00:06:52,606 Which of the following would change the speed 173 00:06:52,606 --> 00:06:53,782 of the sound waves? 174 00:06:53,782 --> 00:06:55,232 Increasing the frequency would just make 175 00:06:55,232 --> 00:06:57,193 that sound a higher note. 176 00:06:57,193 --> 00:06:59,614 But it wouldn't change the speed of the sound wave. 177 00:06:59,614 --> 00:07:01,158 Increasing the temperature of the room 178 00:07:01,158 --> 00:07:04,462 is a change to the medium itself so this would change 179 00:07:04,462 --> 00:07:05,897 the speed of sound. 180 00:07:05,897 --> 00:07:07,945 Decreasing the amplitude's just gonna make the sound 181 00:07:07,945 --> 00:07:10,333 seem softer and not appear as loud. 182 00:07:10,333 --> 00:07:12,898 And decreasing the total volume of the space in the lab room 183 00:07:12,898 --> 00:07:15,064 doesn't actually effect the medium, 184 00:07:15,064 --> 00:07:16,755 it just gives you less of the medium. 185 00:07:16,755 --> 00:07:18,937 So the best answer here would be B. 186 00:07:18,937 --> 00:07:21,919 The doppler effect refers to the change in the perceived 187 00:07:21,919 --> 00:07:24,705 frequency when a speaker or a wave source, 188 00:07:24,705 --> 00:07:26,809 moves relative to the observer. 189 00:07:26,809 --> 00:07:29,825 If a wave source and observer are moving toward each other, 190 00:07:29,825 --> 00:07:32,479 the wavelength of the sound wave is gonna decrease 191 00:07:32,479 --> 00:07:33,977 according to that observer. 192 00:07:33,977 --> 00:07:36,841 Which would make the perceived frequency increase. 193 00:07:36,841 --> 00:07:39,355 This happens since if the source is heading toward you, 194 00:07:39,355 --> 00:07:41,911 as that speaker emits wave pulses, 195 00:07:41,911 --> 00:07:45,004 the speaker moves toward the wave pulse it just emitted 196 00:07:45,004 --> 00:07:47,509 and on this leading edge the crests of the wave 197 00:07:47,509 --> 00:07:48,717 will be closer together. 198 00:07:48,717 --> 00:07:51,859 Since they're closer together, the wavelength is smaller. 199 00:07:51,859 --> 00:07:54,116 And the rate at which these crests are gonna hit 200 00:07:54,116 --> 00:07:56,351 the observer is gonna be higher. 201 00:07:56,351 --> 00:07:58,358 So this observer's gonna hear a higher frequency 202 00:07:58,358 --> 00:08:01,066 than is actually being played by the wave source 203 00:08:01,066 --> 00:08:02,447 when it is at rest. 204 00:08:02,447 --> 00:08:04,422 And for the observer on the trailing edge, 205 00:08:04,422 --> 00:08:06,912 since the wave source is moving away from the pulses 206 00:08:06,912 --> 00:08:08,500 it just sent in this direction, 207 00:08:08,500 --> 00:08:11,008 these wave crests are gonna be spaced further apart, 208 00:08:11,008 --> 00:08:13,612 which increases the wavelength and decreases 209 00:08:13,612 --> 00:08:16,635 the rate at which these crests are gonna hit the observer. 210 00:08:16,635 --> 00:08:18,558 So this observer's gonna hear a frequency 211 00:08:18,558 --> 00:08:21,302 that's less than the actual frequency being played 212 00:08:21,302 --> 00:08:23,421 by the source when it's at rest. 213 00:08:23,421 --> 00:08:24,947 So what's an example problem involving 214 00:08:24,947 --> 00:08:26,682 the Doppler effect look like? 215 00:08:26,682 --> 00:08:28,882 Let's say the driver of a car sees that they're heading 216 00:08:28,882 --> 00:08:32,212 straight toward a person standing still in the crosswalk. 217 00:08:32,212 --> 00:08:34,507 So the driver continuously honks their horn 218 00:08:34,508 --> 00:08:37,152 and emits a sound of frequency f horn. 219 00:08:37,152 --> 00:08:38,876 This is the frequency the horn plays 220 00:08:38,876 --> 00:08:40,578 when the car would be at rest. 221 00:08:40,578 --> 00:08:42,898 And the driver of the car also simultaneously 222 00:08:42,898 --> 00:08:45,378 slams on the brakes and skids to a stop 223 00:08:45,378 --> 00:08:48,046 right in front of the person standing in the crosswalk. 224 00:08:48,046 --> 00:08:49,579 What would that person in the crosswalk 225 00:08:49,579 --> 00:08:52,364 hear as the car's skidding to a stop? 226 00:08:52,364 --> 00:08:54,265 Well since that car is heading toward the person, 227 00:08:54,265 --> 00:08:57,144 those wave crests are gonna be spaced closer together, 228 00:08:57,144 --> 00:08:59,497 so this person will hear a smaller wavelength 229 00:08:59,497 --> 00:09:01,247 and a higher frequency. 230 00:09:01,247 --> 00:09:03,801 But as the car slows down, this effect becomes less 231 00:09:03,801 --> 00:09:04,790 and less dramatic. 232 00:09:04,790 --> 00:09:07,502 And once the car stops, the wave crests will be 233 00:09:07,502 --> 00:09:09,424 spaced out their normal spacing, 234 00:09:09,424 --> 00:09:11,932 and the person will just hear the regular horn 235 00:09:11,932 --> 00:09:13,581 frequency of the car. 236 00:09:13,581 --> 00:09:16,140 So first this person's gonna hear a higher frequency, 237 00:09:16,140 --> 00:09:18,534 but that will eventually just become the actual 238 00:09:18,534 --> 00:09:21,379 frequency of the horn once the car stops 239 00:09:21,379 --> 00:09:23,294 and there's no more relative motion between 240 00:09:23,294 --> 00:09:24,699 the person and the car. 241 00:09:24,699 --> 00:09:27,110 When two waves overlap in the same medium, 242 00:09:27,110 --> 00:09:30,758 we call it wave interference, or wave superposition. 243 00:09:30,758 --> 00:09:32,717 While those waves are overlapping, 244 00:09:32,717 --> 00:09:34,891 they'll combine to form a wave shape 245 00:09:34,891 --> 00:09:37,371 that will be the sum of the two waves. 246 00:09:37,371 --> 00:09:39,466 In other words, while the two waves are overlapping, 247 00:09:39,466 --> 00:09:41,453 to find the value of the total wave, 248 00:09:41,453 --> 00:09:44,955 you just add up the values of the individual waves. 249 00:09:44,955 --> 00:09:47,172 So if you overlap two waves that look identical, 250 00:09:47,172 --> 00:09:50,592 they would combine to form a wave that's twice as big. 251 00:09:50,592 --> 00:09:52,965 And we call this constructive interference. 252 00:09:52,965 --> 00:09:55,719 And if you overlap two waves that are 180 degrees 253 00:09:55,719 --> 00:09:59,805 out of phase, they'll combine to form no wave at all. 254 00:09:59,805 --> 00:10:02,170 We call this destructive interference. 255 00:10:02,170 --> 00:10:04,369 So even though while the waves are overlapping, 256 00:10:04,369 --> 00:10:07,622 you add up their individual values to get the total wave. 257 00:10:07,622 --> 00:10:09,530 After these waves are done overlapping, 258 00:10:09,530 --> 00:10:11,432 they'll pass right through each other. 259 00:10:11,432 --> 00:10:13,133 So if I sent a wave pulse down 260 00:10:13,133 --> 00:10:15,354 a string at another wave pulse, 261 00:10:15,354 --> 00:10:16,954 when these two pulses overlap, 262 00:10:16,954 --> 00:10:19,537 the string would be flat but shortly after that, 263 00:10:19,537 --> 00:10:23,121 the wave pulses would continue on their way unaffected. 264 00:10:23,121 --> 00:10:24,986 They don't bounce off of each other, 265 00:10:24,986 --> 00:10:27,004 or cause permanent damage. 266 00:10:27,004 --> 00:10:28,904 It's only while they're overlapping 267 00:10:28,904 --> 00:10:30,941 that you'll get the wave interference. 268 00:10:30,941 --> 00:10:33,033 So what's an example problem involving wave 269 00:10:33,033 --> 00:10:34,208 interference look like? 270 00:10:34,208 --> 00:10:36,175 Let's say two wave pulses on a string 271 00:10:36,175 --> 00:10:39,412 head toward each other as seen in this diagram to the right 272 00:10:39,412 --> 00:10:41,594 and we wanna know what would be the shape of the wave 273 00:10:41,594 --> 00:10:43,707 when the wave pulses overlap? 274 00:10:43,707 --> 00:10:45,744 So to find the total wave we'll add up the values 275 00:10:45,744 --> 00:10:47,528 of each individual wave. 276 00:10:47,528 --> 00:10:49,467 The blue wave's gonna move to to the right, 277 00:10:49,467 --> 00:10:51,486 and the red wave's gonna move to the left. 278 00:10:51,486 --> 00:10:53,532 And we'll add up the individual values. 279 00:10:53,532 --> 00:10:56,631 Zero of the red wave plus negative two units 280 00:10:56,631 --> 00:10:59,958 of the blue wave will add up to negative two units 281 00:10:59,958 --> 00:11:01,340 for the total wave. 282 00:11:01,340 --> 00:11:03,495 And then positive two units of the red wave 283 00:11:03,495 --> 00:11:06,004 plus negative two units of the blue wave 284 00:11:06,004 --> 00:11:08,848 is gonna equal zero units for the total wave. 285 00:11:08,848 --> 00:11:11,069 Again positive two units of the red wave 286 00:11:11,069 --> 00:11:13,124 plus negative two units of the blue wave 287 00:11:13,124 --> 00:11:15,385 add up to zero units for the total wave. 288 00:11:15,385 --> 00:11:17,238 And then zero units for the red wave 289 00:11:17,238 --> 00:11:19,951 plus negative two units for the blue wave 290 00:11:19,951 --> 00:11:23,135 is gonna equal negative two units for the total wave. 291 00:11:23,135 --> 00:11:25,414 So our total wave will look like this. 292 00:11:25,414 --> 00:11:27,328 Since this pyramid just took a bite out of this 293 00:11:27,328 --> 00:11:29,240 blue rectangular wave. 294 00:11:29,240 --> 00:11:31,918 How do you deal with standing waves on strings? 295 00:11:31,918 --> 00:11:33,541 Well to get a standing wave at all, 296 00:11:33,541 --> 00:11:35,167 you need waves overlapping 297 00:11:35,167 --> 00:11:37,215 that are going in opposite directions. 298 00:11:37,215 --> 00:11:39,231 But even if you have that, you're not necessarily 299 00:11:39,231 --> 00:11:40,717 gonna get a standing wave. 300 00:11:40,717 --> 00:11:42,933 Only certain allowed wavelengths will create 301 00:11:42,933 --> 00:11:45,015 a standing wave in that medium. 302 00:11:45,015 --> 00:11:47,127 And what determines the allowed wave lengths 303 00:11:47,127 --> 00:11:49,722 is the length of the medium and the boundaries 304 00:11:49,722 --> 00:11:50,906 of that medium. 305 00:11:50,906 --> 00:11:53,352 In other words the ends of a string could be fixed 306 00:11:53,352 --> 00:11:56,248 or loose, if the end of the string is fixed, 307 00:11:56,248 --> 00:11:58,555 it's gonna be a displacement node. 308 00:11:58,555 --> 00:12:00,657 Node is the word we use to refer to points 309 00:12:00,657 --> 00:12:02,307 that have no displacement. 310 00:12:02,307 --> 00:12:03,771 And if the end of a string is loose, 311 00:12:03,771 --> 00:12:06,697 that end would act as a displacement anti node. 312 00:12:06,697 --> 00:12:10,273 Anti node being a point that has maximum displacement. 313 00:12:10,273 --> 00:12:11,897 What would these standing waves look like? 314 00:12:11,897 --> 00:12:15,280 Standing waves no longer appear to move along the medium, 315 00:12:15,280 --> 00:12:18,285 they just oscillate back and forth in place. 316 00:12:18,285 --> 00:12:21,913 So this anti node would move from the top to the bottom 317 00:12:21,913 --> 00:12:23,014 back to the top. 318 00:12:23,014 --> 00:12:26,047 But you wouldn't see this crest move right or left, 319 00:12:26,047 --> 00:12:28,109 hence the name standing wave. 320 00:12:28,109 --> 00:12:30,047 However the nodes stay put. 321 00:12:30,047 --> 00:12:32,478 There's never any displacement at a node. 322 00:12:32,478 --> 00:12:34,456 Once you've determined the boundary conditions 323 00:12:34,456 --> 00:12:35,989 and the length of that medium, 324 00:12:35,989 --> 00:12:37,855 the possible wavelengths are set. 325 00:12:37,855 --> 00:12:39,638 Because the only allowed standing waves 326 00:12:39,638 --> 00:12:42,070 have to start at a node and end at a node. 327 00:12:42,070 --> 00:12:44,538 The fundamental standing wave refers to the largest 328 00:12:44,538 --> 00:12:46,708 possible wavelength standing wave. 329 00:12:46,708 --> 00:12:49,661 And in this case it would be one half of a wavelength. 330 00:12:49,661 --> 00:12:52,034 So the length of the medium would have to equal one half 331 00:12:52,034 --> 00:12:54,086 of the wavelength of the wave. 332 00:12:54,086 --> 00:12:56,155 Similarly for the second harmonic, 333 00:12:56,155 --> 00:12:58,303 we still have to start and end at a node, 334 00:12:58,303 --> 00:13:01,264 so the next possibility would be an entire wavelength 335 00:13:01,264 --> 00:13:03,912 which means the length of the medium would equal one 336 00:13:03,912 --> 00:13:05,182 wavelength of the wave. 337 00:13:05,182 --> 00:13:06,633 And if you get to the third harmonic, 338 00:13:06,633 --> 00:13:08,984 this is three halves of a wavelength. 339 00:13:08,984 --> 00:13:11,456 And there's no limit, you can keep going here. 340 00:13:11,456 --> 00:13:13,698 To excite these higher standing waves, 341 00:13:13,698 --> 00:13:15,826 you gotta keep increasing the frequency, 342 00:13:15,826 --> 00:13:18,069 because you'll keep decreasing the wavelength 343 00:13:18,069 --> 00:13:20,057 of the distance between peaks. 344 00:13:20,057 --> 00:13:21,493 So what would a standing wave look like 345 00:13:21,493 --> 00:13:23,435 if one of the ends were loose? 346 00:13:23,435 --> 00:13:26,225 In that case that end would be an anti node, 347 00:13:26,225 --> 00:13:28,569 and the fundamental standing wave would only take the shape 348 00:13:28,569 --> 00:13:30,381 of a fourth of a wavelength. 349 00:13:30,381 --> 00:13:33,227 Since it has to go from a node to an anti node. 350 00:13:33,227 --> 00:13:35,404 That means the length of the string would have to equal 351 00:13:35,404 --> 00:13:36,906 one fourth of a wavelength. 352 00:13:36,906 --> 00:13:38,427 The next possible standing wave would be 353 00:13:38,427 --> 00:13:40,328 three fourths of a wavelength. 354 00:13:40,328 --> 00:13:42,573 And the next possibility would be five fourths 355 00:13:42,573 --> 00:13:43,567 of a wavelength. 356 00:13:43,567 --> 00:13:45,810 And again this progression keeps on going. 357 00:13:45,810 --> 00:13:48,204 So what's an example problem involving standing waves 358 00:13:48,204 --> 00:13:49,583 on strings look like? 359 00:13:49,583 --> 00:13:52,077 Let's say one end of a string of length L 360 00:13:52,077 --> 00:13:54,655 is attached to the wall and the other end is fixed 361 00:13:54,655 --> 00:13:56,266 to a vibrating rod. 362 00:13:56,266 --> 00:13:58,820 A student finds that the string sets up a standing wave 363 00:13:58,820 --> 00:14:00,845 as seen here when the frequency 364 00:14:00,845 --> 00:14:03,167 of the rod is set to f nought. 365 00:14:03,167 --> 00:14:05,895 What are the speed of the waves on the string? 366 00:14:05,895 --> 00:14:07,484 Well we know the string length is L. 367 00:14:07,484 --> 00:14:09,709 And we can figure out how much of a wavelength this is. 368 00:14:09,709 --> 00:14:12,903 From here to here would be one wavelength. 369 00:14:12,903 --> 00:14:14,377 And there's another half. 370 00:14:14,377 --> 00:14:17,144 So the string length L is equalling three halves 371 00:14:17,144 --> 00:14:19,662 of a wavelength or in other words the wavelength here 372 00:14:19,662 --> 00:14:21,497 is 2L over three. 373 00:14:21,497 --> 00:14:23,486 And we know the speed of the wave is always 374 00:14:23,486 --> 00:14:25,486 wavelength times frequency. 375 00:14:25,486 --> 00:14:28,739 So the speed of the wave here is gonna be 2L over three, 376 00:14:28,739 --> 00:14:30,691 which is the wavelength of the wave, 377 00:14:30,691 --> 00:14:33,913 times the frequency and so the best answer is D. 378 00:14:33,913 --> 00:14:36,670 How do you deal with standing waves in tubes? 379 00:14:36,670 --> 00:14:38,615 So just like standing waves in strings, 380 00:14:38,615 --> 00:14:40,785 the wavelength of the standing waves in a tube 381 00:14:40,785 --> 00:14:42,831 are determined by the length of the tube, 382 00:14:42,831 --> 00:14:45,204 and the boundary conditions of that tube. 383 00:14:45,204 --> 00:14:47,235 But this time instead of the sting shaking, 384 00:14:47,235 --> 00:14:50,550 you're creating a standing wave out of sound waves. 385 00:14:50,550 --> 00:14:52,844 Now for the boundary conditions of this tube, 386 00:14:52,844 --> 00:14:56,414 an open end is gonna act like a displacement anti node. 387 00:14:56,414 --> 00:14:59,919 Since the air at an open end can oscillate wildly. 388 00:14:59,919 --> 00:15:02,111 And you get maximum air disturbance. 389 00:15:02,111 --> 00:15:03,924 But a closed end of a tube is gonna act like 390 00:15:03,924 --> 00:15:07,284 a displacement node since there'll be no air disturbance 391 00:15:07,284 --> 00:15:08,582 at a closed end. 392 00:15:08,582 --> 00:15:10,760 So what if both ends of the tube were open 393 00:15:10,760 --> 00:15:13,741 giving us anti node anti node standing waves? 394 00:15:13,741 --> 00:15:16,428 Well the first possibility of the largest wavelength 395 00:15:16,428 --> 00:15:19,802 standing wave would go from anti node to anti node. 396 00:15:19,802 --> 00:15:21,967 And this is one half of a wavelength. 397 00:15:21,967 --> 00:15:23,699 So the length of this tube would have to equal 398 00:15:23,699 --> 00:15:25,276 half of a wavelength. 399 00:15:25,276 --> 00:15:27,715 The next possibility would still go from anti node 400 00:15:27,715 --> 00:15:28,744 to anti node. 401 00:15:28,744 --> 00:15:31,232 And this is equal to one whole wavelength. 402 00:15:31,232 --> 00:15:34,076 It might not look like is but from valley to valley, 403 00:15:34,076 --> 00:15:35,635 is a whole wavelength. 404 00:15:35,635 --> 00:15:37,972 So the length of this tube would equal one wavelength. 405 00:15:37,972 --> 00:15:40,491 And the next possibility would equal three halves 406 00:15:40,491 --> 00:15:41,646 of a wavelength. 407 00:15:41,646 --> 00:15:43,785 And you should note this is the same progression 408 00:15:43,785 --> 00:15:46,592 that we had for node node strings. 409 00:15:46,592 --> 00:15:49,025 So whether both ends are anti nodes, 410 00:15:49,025 --> 00:15:51,337 or both ends are nodes, 411 00:15:51,337 --> 00:15:53,821 if both boundary conditions are the same, 412 00:15:53,821 --> 00:15:55,504 you get this same progression that goes 413 00:15:55,504 --> 00:15:58,246 half of a wavelength, one whole wavelength, 414 00:15:58,246 --> 00:15:59,888 three halves of a wavelength. 415 00:15:59,888 --> 00:16:03,078 Basically any integer or half integer wavelength. 416 00:16:03,078 --> 00:16:05,606 And what if we closed one of the ends of this tube? 417 00:16:05,606 --> 00:16:07,579 If we closed one end of the tube, 418 00:16:07,579 --> 00:16:09,959 that end would become a displacement node, 419 00:16:09,959 --> 00:16:12,170 since the air can't move at that position. 420 00:16:12,170 --> 00:16:14,117 Which would make it a node and it would have to go 421 00:16:14,117 --> 00:16:16,152 to the open end which is an anti node. 422 00:16:16,152 --> 00:16:18,159 So the largest possibility this time would 423 00:16:18,159 --> 00:16:19,846 be one fourth of a wavelength. 424 00:16:19,846 --> 00:16:23,269 The next possibility would still go from node to anti node 425 00:16:23,269 --> 00:16:25,389 and this would be three fourths of a wavelength. 426 00:16:25,389 --> 00:16:27,631 And if you notice, this is exactly the same 427 00:16:27,631 --> 00:16:30,547 as when we had node anti node strings. 428 00:16:30,547 --> 00:16:33,180 We had the same progression of lambda over four, 429 00:16:33,180 --> 00:16:36,227 three lamba over four, five lambda over four, 430 00:16:36,227 --> 00:16:38,856 any odd integer lambda over four, 431 00:16:38,856 --> 00:16:41,306 were the allowed standing wavelengths. 432 00:16:41,306 --> 00:16:43,991 So if one end has a different boundary condition 433 00:16:43,991 --> 00:16:46,148 from the other end this is gonna be the progression 434 00:16:46,148 --> 00:16:48,392 of allowed lengths of the medium. 435 00:16:48,392 --> 00:16:50,685 So what would an example problem involving standing waves 436 00:16:50,685 --> 00:16:52,092 in tubes look like? 437 00:16:52,092 --> 00:16:54,646 Let's say you blow over the top of a tube that's open 438 00:16:54,646 --> 00:16:58,012 at both ends and it resonates with a frequency f nought. 439 00:16:58,012 --> 00:17:00,045 If the bottom of the tube is then covered 440 00:17:00,045 --> 00:17:02,759 and air is again blown over the top of the tube, 441 00:17:02,759 --> 00:17:05,748 what frequency would be heard relative to the frequency 442 00:17:05,748 --> 00:17:07,618 heard when both ends were open? 443 00:17:07,618 --> 00:17:09,071 Well when both ends are open, 444 00:17:09,071 --> 00:17:11,516 we know the standing wave's gonna be an anti node 445 00:17:11,517 --> 00:17:12,762 to anti node. 446 00:17:12,762 --> 00:17:14,811 Which is one half of a wavelength. 447 00:17:14,811 --> 00:17:16,834 Which would equal the length of that tube. 448 00:17:16,835 --> 00:17:19,056 So the wavelength would be two times the length 449 00:17:19,056 --> 00:17:19,983 of the tube. 450 00:17:19,983 --> 00:17:21,842 But when we close one of the ends, 451 00:17:21,843 --> 00:17:25,142 we turn that end from an anti node into a node. 452 00:17:25,142 --> 00:17:27,973 So we'd have to go from an anti node to a node. 453 00:17:27,973 --> 00:17:30,333 Which is only one fourth of a wavelength. 454 00:17:30,333 --> 00:17:31,882 So one fourth of a wavelength would equal 455 00:17:31,882 --> 00:17:33,153 the length of the tube. 456 00:17:33,153 --> 00:17:37,063 And that means lambda equals 4L, this wavelength doubled. 457 00:17:37,063 --> 00:17:38,931 So what would that do to the frequency? 458 00:17:38,931 --> 00:17:40,764 Well we know V equals lambda f, 459 00:17:40,764 --> 00:17:42,580 and we didn't change the medium here 460 00:17:42,580 --> 00:17:44,402 so the speed is gonna remain the same. 461 00:17:44,402 --> 00:17:46,158 So if we double the wavelength, 462 00:17:46,158 --> 00:17:48,114 we'd have to cut the frequency in half 463 00:17:48,114 --> 00:17:51,288 in order to maintain the same speed of the wave. 464 00:17:51,288 --> 00:17:53,312 So when we close the bottom of this tube, 465 00:17:53,312 --> 00:17:55,374 we'd hear half the frequency we heard 466 00:17:55,374 --> 00:17:57,282 when both ends were open. 467 00:17:57,282 --> 00:17:59,262 Beat frequency refers to the phenomenon 468 00:17:59,262 --> 00:18:02,534 where two waves overlap with different frequencies. 469 00:18:02,534 --> 00:18:04,618 When this occurs, the interference of the waves 470 00:18:04,618 --> 00:18:07,701 at a point in space turns from constructive, 471 00:18:07,701 --> 00:18:11,207 to destructive back to constructive and so on. 472 00:18:11,207 --> 00:18:12,824 Which if this were a sound wave, 473 00:18:12,824 --> 00:18:16,226 you'd perceive as a wobble in the loudness of the sound. 474 00:18:16,226 --> 00:18:18,505 And the reason this happens is that if these waves 475 00:18:18,505 --> 00:18:21,264 started in phase and they were constructive, 476 00:18:21,264 --> 00:18:23,222 since they have different frequencies, 477 00:18:23,222 --> 00:18:26,176 one wave would start to become out of phase with the other. 478 00:18:26,176 --> 00:18:29,892 Eventually becoming destructive, which would be soft. 479 00:18:29,892 --> 00:18:31,902 But if you wait longer, one of these peaks 480 00:18:31,902 --> 00:18:34,593 catches up to the next peak in the progression, 481 00:18:34,593 --> 00:18:36,761 and the waves again become constructive 482 00:18:36,761 --> 00:18:38,052 which would be loud again. 483 00:18:38,052 --> 00:18:40,913 And the times this takes to go from loud to soft 484 00:18:40,913 --> 00:18:43,454 to loud again is called the beat period. 485 00:18:43,454 --> 00:18:45,732 But more often you'll hear about the beat frequency. 486 00:18:45,732 --> 00:18:48,329 Which is just one over the beat period. 487 00:18:48,329 --> 00:18:50,594 So the beat period is the time it takes to go from 488 00:18:50,594 --> 00:18:52,647 loud to soft back to loud. 489 00:18:52,647 --> 00:18:54,418 And the beat frequency is the number 490 00:18:54,418 --> 00:18:57,007 of times it does that per second. 491 00:18:57,007 --> 00:19:00,507 How do you determine the beat frequency or the beat period? 492 00:19:00,507 --> 00:19:02,767 Well the formula used to find the beat frequency 493 00:19:02,767 --> 00:19:04,385 is actually really simple. 494 00:19:04,385 --> 00:19:06,344 You just take the difference of the frequencies 495 00:19:06,344 --> 00:19:08,419 of the two waves that are overlapping. 496 00:19:08,419 --> 00:19:09,955 If there is no difference, 497 00:19:09,955 --> 00:19:11,903 if these waves have the same frequency, 498 00:19:11,903 --> 00:19:13,650 you'd have a beat frequency of zero 499 00:19:13,650 --> 00:19:16,118 which would mean you hear no wobbles at all. 500 00:19:16,118 --> 00:19:18,513 The further apart these two frequencies get, 501 00:19:18,513 --> 00:19:20,995 the more wobbles you would hear per second. 502 00:19:20,995 --> 00:19:22,308 And then to find the beat period, 503 00:19:22,308 --> 00:19:24,369 you could take one over the beat frequency. 504 00:19:24,369 --> 00:19:25,924 So what would an example problem 505 00:19:25,924 --> 00:19:27,813 involving beat frequency look like? 506 00:19:27,813 --> 00:19:29,786 Let's say these two waves were overlapping 507 00:19:29,786 --> 00:19:31,767 and we want to determine the beat frequency. 508 00:19:31,767 --> 00:19:34,226 The period of the first wave is four seconds. 509 00:19:34,226 --> 00:19:37,188 That means the frequency of the first wave is one over four, 510 00:19:37,188 --> 00:19:38,355 or 0.25 hertz. 511 00:19:39,626 --> 00:19:41,984 And the period of the second wave is two seconds, 512 00:19:41,984 --> 00:19:46,033 which means that the frequency is one over two or 0.5 hertz. 513 00:19:46,033 --> 00:19:48,948 To get the beat frequency you subtract one frequency 514 00:19:48,948 --> 00:19:49,995 from the other. 515 00:19:49,995 --> 00:00:00,000 0.5 minus 0.25 would be 0.25 hertz.