course Phy 232 ÊõÞªÝÆü‹€ùÿç«»G©©o§£ÓX¥þ¿G®assignment #023
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10:40:14 query doppler shift experiment (experiment was to be read and viewed only) **** explain why the frequency of the sound observed when the buzzer moves toward you is greater than that of the stationary buzzer and why this frequency is greater than that observed when the buzzer is moving away from you
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RESPONSE --> frequency when it moves towards me is greater b/c as each wave starts, it actaully
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10:41:20 ** 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. **
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RESPONSE --> I like yours, it's shorter!
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10:42:49 query General College Physics and Principles of Physics: what is a decibel?
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RESPONSE --> --- skipped sound intensity
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10:43:06 ** 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. **
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RESPONSE -->
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10:55:48 gen 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
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RESPONSE --> air pressure changes along with medium displacement
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10:56:56 ** 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 are NAN, NANAN, NANANAN, ... , containing 1, 2, 3, 4, ..., half-wavelengths in the length of the string. Possible wavelengths are therefore 2 L, 1 L, 2/3 L, ..., where L is string length. For an open organ pipe 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, ... **
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RESPONSE --> not what I got....but I have it now.
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10:58:25 **** gen phy what are beats?
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RESPONSE --> ""beats are fluctuations in amplitude produced by two sound waves of slightly different frequency (16 & 18 Hz)"" superpose one wave onto the other
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10:58:38 ** 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. **
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RESPONSE --> right
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11:02:09 **** query univ phy 16.54 (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?
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RESPONSE --> by holding it the middle, the antinodes at the ends are in phase with each other and can achieve the largest wavelength with the smallest freq. at L = 1.5 m v = 5941 m/s equation 16.18 fn = nv/2L = freq 5941/(2)1.5 = 1980.33 holding one end, doubles the L the wave has available so 1980.33 * 2 = 3960.66
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11:03:29 STUDENT 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. **
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RESPONSE --> good explanation, thank you.
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