#$&* course PHY 202 014. `Query 14 was resubmitted 1 Jul 2011 around 9:00 PM. 014. `Query 14*********************************************
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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 dimension which will be towards and away. The buzzer is actually moving in a circle which means it exists in three dimensions but is moving in two dimensions 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 dimension. 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 perceive 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 perceived 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 therefore 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 perceived 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 perceived by the listener will be lower than the actual frequency. This is because, again at rest the actual frequency will be the perceived 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 perceived frequency than the actual frequency. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery General College Physics and Principles of Physics: what is a decibel? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A decibel is one unit on the decibel scale, which is a logarithmic scale. The name means one-tenth of a bel, a bel being an eponymous unit named for Alexander Graham Bell and used to compare power in electrical communication, voltage, or intensity of sound. The abbreviation of bel is B and decibel, dB. 10 dB = 1 B confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 dimensions 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 perceive 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 perceived 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 determining the intensity in decibels is ... Intensity in decibels = 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 decibels or inaudible. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: 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.. For an open organ pipe, there are nodes at both ends so the configuration must be A &&& A. Possibilities include ANA, ANANA, ANANANA, ANANANANA 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. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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.. 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: 3 ********************************************* Question: `q **** gen phy what are beats? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Beats happens when the two sounds are close in frequency. Beats also occur when the combined sound alternate from louder to quieter then louder etc. with a frequency equal to the differences of the frequencies of the two sounds. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!