#$&* course Phy 202 Sunday, July 15, 2012, between 12:50 PM and 12:55 PM If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: `a** The 'pulses' emitted by an approaching source in a certain time interval are all received in a shorter time interval, since the last 'pulse' is emitted closer to the source than the first and therefore arrives sooner than if the source was still. So the frequency is higher. If the source is moving away then the last 'pulse' is emitted further from the source than if the source was still, hence arrives later, so the pulses are spread out over a longer time interval and the frequency is lower. GOOD EXPLANATION FROM STUDENT: Well, for the purposes of this explanation, I am going to explain the movement of the buzzer in one dimention which will be towards and away. The buzzer is actually moving in a circle which means it exists in three dimentions but is moving in two dimentions with relation to the listener. However, using trigonometry we can determine that at almost all times the buzzer is moving either towards or away from the listener so I will explain this in terms of one dimention. }When a buzzer is 'buzzing' it is emitting sound waves at a certain frequency. This frequency appears to change when the buzzer moves toward or away from the listener but the actual frequency never changes from the original frequency. By frequency we mean that a certain number of sound waves are emitted in a given time interval (usually x number of cycles in a second). So since each of the waves travel at the same velocity they will arrive at a certain vantage point at the same frequency that they are emitted. So If a 'listener' were at this given vantage point 'listening', then the listener would percieve the frequency to be what it actually is. Now, if the buzzer were moving toward the listener then the actual frequency being emitted by the buzzer would remain the same. However, the frequency percieved by the listener would be higher than the actual frequency. This is because, at rest or when the buzzer is not moving, all of the waves that are emitted are traveling at the same velocity and are emitted from the same location so they all travel the same distance. But, when the buzzer is moving toward the listener, the waves are still emitted at the same frequency, and the waves still travel at the same velocity, but the buzzer is moving toward the listener, so when a wave is emitted the buzzer closes the distance between it and the listener a little bit and there fore the next wave emitted travels less distance than the previous wave. So the end result is that each wave takes less time to reach the listener than the previously emitted wave. This means that more waves will reach the listener in a given time interval than when the buzzer was at rest even though the waves are still being emitted at the same rate. This is why the frequency is percieved to be higher when the buzzer is moving toward the listener. By the same token, if the same buzzer were moving away from the listener then the actual frequency of the waves emitted from the buzzer would be the same as if it were at rest, but the frequency percieved by the listener will be lower than the actual frequency. This is because, again at rest the actual frequency will be the percieved frequency. But when the buzzer is moving away from the listener, the actual frequency stays the same, the velocity of the waves stays the same, but because the buzzer moves away from the listener a little bit more each time it emits a wave, the distance that each wave must travel is a little bit more than the previously emitted wave. So therefore, less waves will pass by the listener in a given time interval than if the buzzer were not moving. This will result in a lower percieved frequency than the actual frequency. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK (I considered going into the explanation about how a bug standing on the buzzer would always hear the same frequency, etc., but I thought that that would be redundant/overkill.) ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qquery General College Physics and Principles of Physics: what is a decibel? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - SOLUTION: A decibel is a fixed ratio (not exactly a ""unit"" or ""increment"") of sound intensity between two sounds, such that a) a sound with intensity one decibel greater than another's has intensity equal to the other's intensity times (1 + the tenth root of ten), such that an increment of +10 dB (one ""bel"") represents a tenfold increase in sound intensity, and b) a sound having 0 dB is at the average threshold of human hearing. Thus, a sound-intensity measurement in decibels represents 10 * (the power of 10 by which the intensity is a multiple of the intensity at the threshold of human hearing), i.e., 10 * the ten-base logarithm of that multiple. confidence rating #$&*:e: how well I understand it; varying between 3 and 2 re: my ability to express it ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** dB = 10 log( I / I0 ), where I is the intensity of the sound in units of power per unit area and I0 is the 'hearing threshold' intensity. MORE EXTENSIVE EXPLANATION FROM STUDENT: Sound is possible because we exist in a medium of air. When a sound is emitted, a concussive force displaces the air around it and some amount energy is transferred into kinetic energy as air particles are smacked away from the force. These particles are now moving away from the initial force and collide into other air particles and send them moving and ultimately through a series of collisions the kinetic energy is traveling out in all directions and the air particles are what is carrying it. The behavior of this kinetic energy is to travel in waves. These waves each carry some amount of kinetic energy and the amount of energy that they carry is the intensity of the waves. Intensities of waves are given as a unit of power which is watts per square meter. Or since the waves travel in all directions they move in three dimentions and this unit measures how many watts of energy hits a square meter of the surface which is measuring the intensity. But we as humans don't percieve the intensities of sound as they really are. For example, a human ear would percieve sound B to be twice as loud as sound A when sound B is actually 10 times as loud as sound A. Or a sound that is ... 1.0 * 10^-10 W/m^2 is actually 10 times louder than a sound that is 1.0 * 10^-11 W/m^2 but the human ear would percieve it to only be twice as loud. The decibel is a unit of intensity for sound that measures the intensity in terms of how it is percieved to the human ear. Alexander Graham Bell invented the decibel. Bell originally invented the bel which is also a unit of intensity for waves. The decibel is one tenth of a bel and is more commonly used. The formula for determing the intensity in decibels is ... Intensity in decibles = the logarithm to the base 10 of the sound's intensity/ I base 0 I base 0 is the intensity of some reference level and is usually taken as the minimum intensity audible to an average person which is also called the 'threshold of hearing'. Since the threshold of hearing is in the denominator, if a sound is this low or lower the resulting intensity will be 0 decibles or inaudible. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qgen phy what is the difference between the node-antinode structure of the harmonics a standing wave in a string and in an organ pipe closed at one end YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - SOLUTION: The organ pipe closed at one end produces odd-numbered harmonics / even-numbered overtones only, whereas the string produces both odd- and even-numbered harmonics (this is why a stringed instrument sounds ""warmer"" than a flute or pipe organ): A closed tube has to have a pressure node at the open end so that the wave will be stable with respect to the outside environment rather than dissipating into it (and thereby failing to ""stand""). It must likewise have a displacement node at the closed end because the air there cannot move farther out, and it must therefore have a pressure antinode at the closed end. Having a pressure antinode at the closed end, especially when there is a pressure node at the open one, is impossible for an even-numbered harmonic (a/k/a an odd-numbered overtone), given that even-numbered harmonics are bilaterally symmetrical about their wavepath and, more specifically, have pressure nodes at both ends. Therefore, an even-numbered harmonic or odd-numbered overtone cannot exist as a standing wave in a closed tube.
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Given Solution: `a** In a string there are nodes at both ends so the harmonics are described the the configurations NAN, NANAN, NANANAN, etc.. In a pipe closed at one end there is a node at one end and an antinode at the other so the possible configurations are NA, NANA, NANANA, etc.. ???? Does it follow that one can play ""Hey Jude"" only on closed pipes? ???? displacement nodes are at both ends of the string, so the structure is N &&& N, where &&& is any sequence of nodes and antinodes that results in an alternating sequence. * The possibilities for the fixed-end string are therefore NAN, NANAN, NANANAN, ... , containing 2, 4, 6, 8, ..., quarter-wavelengths in the length of the string. * Possible wavelengths are therefore 2 L, 1 L, 2/3 L, ..., where L is the length of the string. For an open organ pipe, there are nodes at both ends so the configuration must be A &&& A. * Possibilities include ANA, ANANA, ANANANA, ANANANANA, ..., containing 2, 4, 6, 8, ..., quarter-wavelengths. * Possible wavelengths are therefore 2 L, 1 L, 2/3 L, ..., where L is the length of the pipe. * These possible wavelengths are the same as for a fixed-end string of the same length. For an organ pipe open at one end and closed at the other, the configuration must be N &&& A. * Possibilities include NA, NANA, NANANA, NANANANA, ..., containing 1, 3, 5, 7, ..., quarter-wavelengths. * Possible wavelengths are therefore 4 L, 4/3 L, 4/5 L, 4/7 L, ... STUDENT QUESTION My understanding is that open tube produces all harmonics? INSTRUCTOR RESPONSE When I read over it I decided the given solution should be improved; I've inserted the new solution above. It should be somewhat clearer than the old solution. I think I know, but I'm not 100% sure what you mean by 'all harmonics'. So be sure to ask if my response doesn't answer your question. The open pipe produces only the harmonics that occur with a sequence of nodes and antinodes which includes antinodes at both ends. The wavelengths are the same as for a string of the same length, having nodes at both ends. The closed pipe produces only the harmonics which have a node at the closed end and an antinode at the open end. The resulting sequence of possible wavelengths is therefore different than for an open pipe.< &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q **** gen phy what are beats? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - SOLUTION: ""Beats"" are instances of concussion that occur between sound waves at any intervals other than octave multiples (2^n: n E {integers}). Speaking at the most general level, the ""concussion"" occurs when the waves are ""doing different things,"" i.e., when they are at different points on the node-antinode continuum at the same place in space (e.g., the listener's eardrum) and their waveforms ""cross."" This concussion is especially strong when the pressure nodes of one wave do not exactly match the pressure antinodes of the other wave, because destructive interference is less complete and constructive interference is greater (the waves don't ""cancel each other out"" as much) and there is therefore more ""back and forth"" motion created by the combination of the waves. There are no ""beats"" at octave-multiple intervals because the oscillation waveform at the lower frequency is always ""outside"" the oscillation waveform at the higher frequency; the waveforms never cross, and every node of the lower frequency is also a node of the higher one. Other than that, there will always be ""beats,"" and the higher the numbers in the fully reduced frequency ratio, the more cycles must pass before a node-node match. For example, the nodes of sound waves a perfect fifth apart [i.e., of frequency ratio 3:2], will coincide every 2 cycles of the note with lower pitch / longer wavelength or 3 cycles of the note with higher pitch / shorter wavelength, whereas the nodes of waves a minor third apart [i.e., of frequency ratio 6:5] will coincide every 5 cycles of the note with longer wavelength / lower frequency or every 6 cycles of the note with shorter wavelength / higher frequency. It follows that the closer the frequencies of a pair of notes get to each other, the more they will usually tend to ""beat,"" because unless the frequencies happen to have common factors, it will be a long time before they reach a common multiple. confidence rating #$&*: to 2.75 (Although I understand and can apply all of the above information, I may be missing more in-depth information about some part of the topic.) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Beats are what happens when the two sounds are close in frequency. Beats occur when the combined sound gets louder then quieter then louder etc. with a frequency equal to the differences of the frequencies of the two sounds. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 2.5 to 2.75 (I should have devoted more attention to explaining when / how often and in what timing pattern beats occur; it would be nice to review the explanation of why this is so.) ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q **** query univ phy 16.66 / 16.62 11th edition16.54 (20.32 10th edition) steel rod 1.5 m why hold only at middle to get fund? freq of fund? freq of 1st overtone and where held? **** why can the rod be held only at middle to get the fundamental? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [Omitted as marked for only University Physics students] confidence rating #$&*:A] ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT SOLUTION WITH INSTRUCTOR COMMENTS: Since the rod is a rigid body it can form the smallest frequency corresponding to the largest wavelength only when held at the center. The smallest frequency corresponding to the largest wavelength is known as the fundamental. (a) the fingers act as a sort of pivot and the bar performs a teeter - totter motion on either side of the pivot **note: the motion is more like the flapping of a bird's wings, though it obviously doesn't result from the same sort of muscular action but rather from an elastic response to a disturbance, and the motion attenuates over time**. With this manner of motion, the ends have the greatest amount of amplitude, thus making them the antinodes. (b)Holding the rod at any point other than the center changes the wavelngth of the first harmonic causing one end of the rod to have a smaller wave motion than the other. Since the fundamental frequency is in essence the smallest frequency corresponding to the largest wavelength, offsetting Lwould produce an unappropriatley sized wavelength. (c)Fundamental frequency of a steel rod: ( 5941 m /s) / [(2/1)(1.5m) = 1980.33Hz (d) v / (2/2) L = 3960.67Hz --achieved by holding the rod on one end. INSTRUCTOR COMMENTS: The ends of the rod are free so it can vibrate in any mode for which the ends are antinodes. The greatest possible wavelength is the one for which the ends are antinodes and the middle is a node (form ANA). Since any place the rod is held must be a node (your hand would quickly absorb the energy of the vibration if there was any wave-associated particle motion at that point) this ANA configuration is possible only if you hold the rod at the middle. Since a node-antinode distance corresponds to 1/4 wavelength the ANA configuration consists of 1/2 wavelength. So 1/2 wavelength is 1.5 m and the wavelength is 3 m. At propagation velocity 5941 m/s the 3 m wavelength implies a frequency of 5941 m/s / (3 m) = 1980 cycles/sec or 1980 Hz, approx.. The first overtone occurs with antinodes at the ends and node-antinode-node between, so the configuration is ANANA. This corresponds to 4 quarter-wavelengths, or a full wavelength. The resulting wavelength is 1.5 m and the frequency is about 3760 Hz. THE FOLLOWING INFORMATION MIGHT OR MIGHT NOT BE RELATED TO THIS PROBLEM: STUDENT RESPONSES WITH INSTRUCTOR COMMENTS INSERTED To find the amplitudes at 40, 20 and 10 cm from the left end: The amplitudes are: at 40 cm 0 at 20 cm .004m at 10 cm .002828 m I obtained my results by using the information in the problem to write the equation of the standing wave. Since the cosine function is maximum at 0, I substituted t=0 into the equation and the value of x that I wanted to find the amplitude for. ** wavelength = 192 m/2 / (240 Hz) = .8 m. Amplitude at x is .4 cm sin(2 `pi x / .8 m) **the maximum transverse velocity and acceleration at each of these points are found from the equation of motion: `omega = 2 `pi rad * 240 cycles / sec = 480 `pi rad/s. .004 m * 480 `pi rad/s = 6 m/s approx and .004 m * ( 480 `pi rad/s)^2 = 9000 m/s^2, approx. .0028 m * 480 `pi rad/s = 4 m/s approx and .0028 m * ( 480 `pi rad/s)^2 = 6432 m/s^2, approx. ** STUDENT COMMENT The last student response was very interesting. If I understand it correctly, she took the standing wave equation and found the max by setting the derivative equal to zero. INSTRUCTOR RESPONSE That is a very good statement of the reasoning illustrated in the student's solution. The time derivative of the position function is the velocity function for the particles of material, and if the particles at the point where you hold the bar isn't zero, energy will be lost at that point very quickly, causing the standing wave to dissipate its energy (or to retain too little energy to to form in the first place). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): [N/A] ------------------------------------------------ Self-critique Rating: [N/A]" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ________________________________________ #$&*