WAVES SUPERPOSITION OF WAVES STANDING WAVES RESONANCE

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1 WAVES 1.) A wae is a disturbance that moes through a medium. You can t hae water waes without water!) 2.) A wae produced by a force that is perpendicular to the direction of propagation of the wae disturbance is called a transerse wae. a.) An example of a transerse wae is: i.) the wae produced when a taught string it jiggled. i.) the wae produced when a rock is thrown into a pond. SUPERPOSITION OF WAVES 1.) If waes from two sources moing in the same medium superimpose on one another, the waes will oerlay one another. a.) If the oerlay produces an additie effect, the superposition is said to be constructie superposition. b.) If the oerlay produces a subtractie effect, the superposition is said to be destructie superposition. 3.) A wae produced by a force that is in the same direction as the propagating wae disturbance is called a longitudinal wae. a.) An example of a longitudinal wae is sound wae. 1.) 3.) RESONANCE STANDING WAVES 1.) A spring has one frequency at which it will ibrate. That frequency is called the natural frequency of the spring system. 2.) Apply a force to the spring. If the frequency of the force matches the natural frequency of the spring system, the amplitude of the spring s oscillation will become bigger and bigger. 3.) This is called resonance. 4.) In fact, if you apply a periodic force to system and the frequency of that force matches one of the natural frequencies of the system, the amplitude of the system s oscillation will get bigger and bigger and you will hae a resonance between the force and the system. A standing wae is generated when waes generated by two different sources in a system superimpose creating a single, orderly wae. The classic example of the production of standing waes uses a rope tied rigidly to a fixed structure at both ends. If one end is jiggled so as to generate a wae, that wae will proceed down the rope let s say it moes to the left) until it strikes the fixed end whereupon it will bounce and moe back toward the right. If the jiggle is periodic producing wae after wae, the new waes moing to the left will superimpose with the bounced waes moing to the right. Normally, this superposition will appear chaotic with the rope jumping around indeterminately. But under the right condition, the resulting motion will be ery orderly. 2.) 4.)

2 STANDING WAVES on a ROPE The right condition is essentially that of resonance: If the frequency of the force producing the ibration just happens to match one of the natural frequencies of the oscillating system, the resulting superposition will produce standing waes. The trick is determine what frequency or frequencies) that is. To do that, one has to look at the geometry of the system. 3.) That means that if we know the wae elocity, and if we can determine the appropriate waelength for a gien situation, we can determine the frequency of the wae and, hence, the frequency our force must ibrate at to generate a standing wae. 4.) So how to get the appropriate waelength for a gien system? Let s look at our rope situation. What constraint must our appropriate waelength satisfy? Well, it better hae a node a fixed point) at each end, as each end is tied to a rigid structure and can t moe. What kind of wae will do that? Three are shown below. To see how this works, we ll look at our rope example more closely. But before we do, there are a few obseration we need to make. NOTICE: In all three cases, the end-point constraints are met that is, they ALL hae nodes at both ends of their respectie waelengths). 5.) 7.) 1.) As you know, a wae is traditionally characterized by its frequency!, its waelength! and its wae elocity.! meters/cycle) meters/second) 5.) This is all fine and dandy, but how does it help with our problem? It helps because we know that the span between the ends if a fixed distance L. All we hae to do it link L to the waelengths iewed, and we hae the waelengths in terms of a know quantity. Specifically: a.) For the first situation: We know L, so the question is, How many waelengths are in L? L! cycles passing by per second) 2.) The relationship between these three parameters is:! ". Looking at the wae, we can see that there are two quarter-waelengths ) in L sounds obscure you ll get used to it). That is, we can write:! 4 ) "! 2L. L 2! 4 6.) 8.)

3 6.) Assuming we know the wae elocity this would normally be gien), we can write:! " 2L)" # " 2L 7.) So let s say the wae elocity is 3 m/s and the length of the rope is 2 meters. That means:! 2L ) ) 3 m/s 2 2 m "!.75 sec #1 this unit is the same as a Hertz, Hz) So in general, what is the procedure to be followed in these kinds of problems? They are: a.) Identify the end-point constraints. In our example, it was that there had to be nodes fixed points at each end.) b.) On a sine wae, identify what the waeform is going to look like. You ll see examples of this in class.) c.) Once you know what the waeform should look like, be sure that any internal constraints imposed on the system are met by the waeform. Again, you ll see examples of this in class.) d.) When satisfied, ask the question, How many quarter-waelength are there in L? Put a little differently, fill in the? in the expression: ) L?x! 4 e.) Sole for! in terms of L, then use! " to get the required 9.) 11.) 8.) Apparently, if we jiggle the rope at.75 Hz, we will get a standing wae on the rope that, oer time, looks like: 9.) We could do a similar bit of analysis for the other two waeforms. 10.) ONE OTHER THING: If there had been any internal constraints if, for instance, we had wanted to pinch the rope at L/2 making that point a node, then our waeform wouldn t hae worked look at it there is an anti-node an extreme at L/2) and we would hae had to hae done a bit more thinking you ll see examples of this in class). Pipe Closed at One End Another outstanding example of a standing wae is the waeform that is generated when air is piped through a tube. 1.) Consider a pipe of length L closed at one end. What frequency of sound will stand in the pipe? 2.) In a problem like this, the first thing you hae to do is identity what standing waes will fit in the pipe. To do that, you hae to begin by identifying the end-point constraints. a.) For a pipe closed at one end, the endpoint constraints dictate an anti-node at the open end and a node at the closed end. 10.) 12.)

4 3.) The waeforms that fit the bill are shown below, then reproduced in the ertical: node anti-node anti-node anti-node 6.) We know the speed of sound in air is approximately 330 m/s and we know the relationship between a wae s elocity and its waelength and frequency is! 1 " 1. Assuming the tube s length is 2 meters and ignoring the radius correction mentioned at the bottom of the preious page), we can write:! 1 " 1 2 nd waeform 1 st waeform 3 rd waeform! 1 4L 330 m/s) 42 m) Hz 7.) Put a Hz tuning fork at the mouth of our tube and it will howl quite loudly. 13.) 15.) 4.) Each section of wae has a numerical length equal to the length of the tube, or L. 5.) If we ask the question, How many waelengths are there in L? That is, we need to complete the?! L By examination, there is a quarter waelength in L, so we can write: 1 4! 1 L "! 1 4L Minor Note: In real life, the effectie length of the tube has to be altered due to perturbation effects at the ends. In the case of a singly open tube, the effectie length of the tube isn t L but rather L+.4d, where d is the tube s diameter. For doubly open tube, it s L+.8d. 14.) 8.) Doing the same calculation for the second situation where there is three-quarter of a wae in L, we can write:! 2 " 2! )L 3)330 m/s) 42 m) Hz 9.) Put a Hz tuning fork at the mouth of our tube and it will howl quite loudly. 16.)

5 10.) Doing the same calculation for the third situation where there is fie-quarters of a wae in L, we can write:! 3 " 3! )L 5)330 m/s) 42 m) Hz 14.) On the sine wae presented below, you can see the waeforms that satisfy the endpoint constraints. Once determined, they can be put on the sketch to the right though on a test you probably won t redraw the sketch to fit the actual pipe). anti-node anti-node anti-node anti-node 2 nd waeform 11.) Put a Hz tuning fork at the mouth of our tube and it will howl quite loudly. 17.) 1 st waeform 3 rd waeform 19.) Pipe Open at Both Ends 12.) Consider a pipe of length L open at both ends. What frequency of sound will stand in the pipe? 13.) In a problem like this, the first thing you hae to do is identity what standing waes will fit in the pipe. To do that, you hae to begin by identifying the end-point constraints. a.) For a pipe open at both ends, the endpoint constraints dictate anti-nodes at the both ends. 15.) The lowest frequency--the longest waelength--is shown as the first sketch 1 to the right. Noting that, we can 2! 1 L write:! 1 " 1! )L 2)330 m/s) 2 m) 330 Hz 16.) Send a 330 Hz sound wae down the tube, and it will howl at that 18.) 20.)

6 ! 2 16.) The second lowest frequency--the second longest waelength--is shown as the middle sketch to the right. Noting that! 2 L, we can write:! 2 " 2 L 330 m/s) 2 m) 165 Hz 16.) Send a 165 Hz sound wae down the tube, and it will howl at that 21.) 16.) The second lowest frequency--the second longest waelength--is shown as the middle sketch to the right. Noting 6 that, we can write: 4! 3 3 2! L 3! 3 " 3! 3 $ 3 % & 2 L ' ) 2)330 m/s) 3)2 m) 110 Hz 16.) Send a 110 Hz sound wae down the tube and it will howl at that 15.)