course PHY 202 I submited querys 8,9 and 10 but there weren't any comments. Do you want me to resubmit them?
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20:31:04 Query introductory set six, problems 15-18 how do we determine the energy of a standing wave given the amplitude and frequency of the wave and the mass of the string?
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RESPONSE --> Given the amplitde, frequency, and mass of the string we determine the energy of a standing wave by using the formula: KE = 2 `pi^2 m f^2 A^2
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20:32:26 STUDENT ANSWER AND INSTRUCTOR RESPONSE: Energy = 2*pi^2*m*f^2*A^2 INSTRUCTOR RESPONSE: ** You should understand the way we obtain this formula. We assume that every point of the string in in SHM with amplitude A and frequency f. Since the total energy in SHM is the same as the maximum potential or the max kinetic energy, all we need to do is calculate the max potential energy or kinetic energy of each point on the string and add up the results. Since we know mass, frequency and amplitude, we see that we can calulate the max kinetic energy we can get the result we desire. Going back to the circular model, we see that frequency f and amplitude A imply reference point speed = circumference / period = circumference * frequency = 2 `pi A f. The oscillator at its maximum speed will match the speed of the reference point, so the maximum KE is .5 m v^2 = .5 m (2 `pi A f)^2 = 2 `pi^2 m f^2 A^2. **
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RESPONSE --> ok
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20:34:04 If the ends of two strings are driven in phase by a single simple harmonic oscillator, and if the wave velocities in the strings are identical, but the length of one string exceeds that of the other by a known amount, then how do we determine whether a given frequency will cause the 'far ends' of the strings to oscillate in phase?
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RESPONSE --> The longer length string will be behind if they are moving at the same velocity and frequency. If they are moving at the same speed it simply takes it longer to travel a longer distance. The ends will oscillate in phase if the time difference is the same as a certain number of periods the ends will oscillate in phase.
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20:37:26 ** the question here is whether the far ends of the strings are at the same phase of motion, which occurs only if their lengths differ by exactly one, two, three, ... wavelengths. So we need to find the wavelength corresponding to the given frequency, which need not be a harmonic frequency. Any frequency will give us a wavelength; any wavelength can be divided into the difference in string lengths to determine whether the extra length is an integer number of wavelengths. Alternatively, the pulse in the longer string will be 'behind' the pulse in the shorter by the time required to travel the extra length. If we know the frequency we can determine whether this 'time difference' corresponds to a whole number of periods; if so the ends will oscillate in phase **
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RESPONSE --> ok, which occurs only if their lengths differ by exactly one, two, three, ... wavelengths. So we need to find the wavelength corresponding to the given frequency, which need not be a harmonic frequency.
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20:46:13 General College Physics and Principles of Physics 11.38: AM 550-1600 kHz, FM 88-108 mHz. What are the wavelength ranges?
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RESPONSE --> We use the formula V= Frequency * wavelength 3.00*10^8=wavelength*550=5.4*10^5 3.00*10^8=wavelength*1600=1.88105 3.00*10^8=wavelength*88=3.4*10^6 3.00*10^8=wavelength*108=2.7*10^6
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20:47:58 At 3 * 10^8 m/s: a frequency of 550 kHz = 550 * 10^3 Hz = 5.5 * 10^5 Hz will correspond to a wavelength of 3 * 10^8 m/s / (5.5 * 10^5 cycles / sec) = 545 meters. a frequency of 1600 kHz = 1.6* 10^6 Hz will correspond to a wavelength of 3 * 10^8 m/s / (1.6 * 10^6 cycles / sec) =187 meters. The wavelengths for the FM range are calculated similarly. a frequency of 88.0 mHz= 88.0 * 10^6 Hz = 8.80 * 10^7 Hz will correspond to a wavelength of 3 * 10^8 m/s / (8.80 * 10^7 cycles / sec) = 3.41 meters. The 108 mHz frequency is calculated similarly and corresponds to a wavelength of 2.78 meters.
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RESPONSE --> I failed to convert khz to Hz and mhz to Hz it should have been 3.00*10^8=wavelength*5.50810^5=545 3.00*10^8=wavelength*1.600^6=188 3.00*10^8=wavelength*8.8*10^7=341 3.00*10^8=wavelength*1.08*10^8=278
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20:50:48 General College Physics and Principles of Physics 11.52: What are the possible frequencies of a violin string whose fundamental mode vibrates at 440 Hz?
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RESPONSE --> We know the harmonics are 2,3, and 4. WE simply muliply 440 by each: 2*440= 880 Hz 3*440= 1320 Hz 4*440= 1760 Hz
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20:52:25 The fundamental mode for a string fixed at both ends fits half a wavelength onto the string and therefore has a wavelength equal to double its length. The next three harmonics fit 2, 3 and 4 half-wavelengths into the length of the string and so have respectively 2, 3 and 4 times the frequency of the fundamental. So the first 4 harmonics are fundamental frequency = 440 Hz First overtone or second harmonic frequency = 2 * 440 Hz = 880 Hz Second overtone or third harmonic frequency = 3 * 440 Hz = 1320 Hz Third overtone or fourth harmonic frequency = 4 * 440 Hz = 1760 Hz
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RESPONSE --> ok, the wavelength onto the string and therefore has a wavelength equal to double its length then we would have 2, 3, and 4
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20:55:40 General College Physics Problem: Earthquake intensity is 2.0 * 10^6 J / (m^2 s) at 48 km from the source. What is the intensity at 1 km from the source?
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RESPONSE --> Since the intensity is 2.0 * 10^6 J / (m^2 s) at 48 km from the source one could simply 48/1 to get the intensity. We then multiply it by 2.0*10^6J9m^2s). With that we get: 48/1=2300 2.0 * 10^6 J / (m^2 s)*2300=4.6*10^9J/(m^2s)
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20:56:56 The wave is assumed spherical so its surface area increases as the square of its distance and its intensity, which is power / surface area, decreases as the square of the distance. So the intensity at 1 km will be (48 km / 1 km)^2 = 2300 times as great, or 2300 * 2.0 * 10^6 J / (m^2 s) = 4.6 * 10^9 J/(m^2 s).
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RESPONSE --> ok
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20:58:37 At what rate did energy pass through a 5.0 m^2 area at the 1 km distance?
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RESPONSE --> For this we simply multiply the last answer by 5.0m^2. So we get 4.6*10^9*5.0m^2=2.3*10^10 J/S
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21:00:15 Through a 5 m^2 area the rate of energy passage is therefore 4.6 * 10^9 J / (m^2 s) * 5.0 m^2 = 2.3 * 10^10 J / s, or 23 billion watts.
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RESPONSE --> ok, the answer should be in watts
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~ȶBmgZɗϡ⦗~j̩ assignment #012 rYɐZ Physics II 04-02-2006
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22:01:56 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 --> The reason is that when it is moving closer the pulses are received at shorter time intervals. Thus, the frequency increases. Moving away the pulses take longer to arrive and their is more time between pulses decresing the frequency. A good exampled would be using a tennis ball machine. The closer it comes the quicker the balls come. They come at a higher frequency. The further away the machine goes the longer it takes for the balls to get there decreasing the frequency.
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22:04:02 ** 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 --> ok
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22:05:20 query General College Physics and Principles of Physics: what is a decibel?
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RESPONSE --> The decibel is used to measure sound level. It is a logarithmic unit used to describe a ratio. It is db= 10log(I/Io) where I is the intensity and Io is the intensity of a chosen reference level.
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22:05:35 ** 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 --> ** ok
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22:08:03 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 --> One characteristic of every standing wave pattern is that there are points along the medium which appear to be standing still. These points, sometimes described as points of no displacement, are referred to as nodes. There are other points along the wave which undergo vibrations between a large positive and and large negative displacement. These are the points which undergo the maximum displacement during each vibrational cycle of the standing wave. In a sense, these points are the opposite of nodes, and so they are called antinodes. If one end was tied and the other was not it would be alternating Node Antinode Node and Antinode. Whatever the length it would start on node and end on antinode.
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22:09:34 ** 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 --> ok
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22:11:23 **** gen phy what are beats?
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RESPONSE --> This is a type of interference when two sounds are very close in frequency, but not exactly the same. The regularly spaced intsnsity changes are called beats.
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22:13:48 ** 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 --> ok, they occur when the pases get in and out of phase due to their different wavelengths.
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22:13:54 **** 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 --> ok
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22:13:58 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 --> ok
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⸱W||mщC֢v_ assignment #013 rYɐZ Physics II 04-02-2006
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22:23:46 query experiment to be viewed and read but not performed: transverse and longitudinal waves in aluminum rod what is the evidence that the higher-pitched waves are longitudinal while the lower-pitched waves are transverse?
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RESPONSE --> the higher pitch are longitudinal because of the way they were struck. If you strike it on top on the lenght wise side it will make longitudinal. the lower pitched are transverse. They were struck on the ends. Thus, emminating out.
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22:24:21 STUDENT RESPONSE: The logitudinal waves had a higher velocity. That doesn't provide evidence that the high-pitched wave was longitudinal, since we didn't directly measure the velocity of those waves. The higher-pitches waves were damped out much more rapidly by touching the very end of the rod, along its central axis, than by touching the rod at the end but on the side. The frequency with which pulses arrive at the ear determines the pitch. The amplitude of the wave affects its intensity, or energy per unit area. For a given pitch the energy falling per unit area is proportional to the square of the amplitude. Intensity is also proportional to the square of the frequency. **
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RESPONSE --> ok
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22:27:51 Query General College Physics and Principles of Physics 12.08: Compare the intensity of sound at 120 dB with that of a whisper at 20 dB.
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RESPONSE --> FOr 120db the intensity is 1. This was solved by I=Io*10^Beta/10 So 1.0*10^-12(10^120/10)=1watts FOr the 20db we use the same formula 1.0*10^-12(10^2)=10^-10watts
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22:38:01 The intensity at 120 dB is found by solving the equation dB = 10 log(I / I_threshold) for I. We get log(I / I_threshold) = dB / 10, so that I / I_threshold = 10^(120 / 10) = 12and I = I_threshold * 10^12. Since I_threshold = 10^-12 watts / m^2, we have for dB = 120: I = 10^-12 watts / m^2 * 10^12 = 1 watt / m^2. The same process tells us that for dB = 20 watts, I = I_threshold * 10^(20 / 10) = 10^-12 watts / m^2 * 10^2 = 10^-10 watts / m^2. Dividing 1 watt / m^2 by 10^-10 watts / m^2, we find that the 120 dB sound is 10^10 times as intense, or 10 billion times as intense. A more elegant solution uses the fact that dB_1 - dB_2 = 10 log(I_1 / I_threshold) - ( 10 log(I_2 / I_threshold) ) = 10 log(I_1 / I_threshold) - ( 10 log(I_2 / I_threshold) ) = 10 {log(I_1) - log( I_threshold) - [ ( log(I_2) - log(I_threshold) ]} = 10 { log(I_1) - log(I_2)} = 10 log(I_1 / I_2). So we have 120 - 20 = 100 = 10 log(I_1 / I_2) and log(I_1 / I_2) = 100 / 10 = 10 so that I_1 / I_2 = 10^10.
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RESPONSE --> ok
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22:48:00 Query gen phy 12.30 length of open pipe, 262 Hz at 21 C? **** gen phy What is the length of the pipe?
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RESPONSE --> For the a part we use the formula L=v/2L First we have to find velocity. We can do this by v=(331+.6*21)= 343.6 m/s So we get 262=343.6/2(262)=.655meters Since the pipe is open there are antinodes at both ends of the wave. I had couldn't find how to do it for outside air in part C.
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22:56:46 GOOD STUDENT SOLUTION First we must determine the velocity of the sound waves given the air temperature. We do this using this formula v = (331 + 0.60 * Temp.) m/s So v = (331 + 0.60 * 21) m/s v = 343.6 m/s The wavelength of the sound is wavelength = v / f = 343.6 m/s / (262 Hz) = 0.33 meters. So 262 Hz = 343.6 m/s / 4 * Length Length = 0.33 meters f = v / (wavelength) 262 Hz = [343 m/s] / (wavelength) wavelength = 1.3 m. So the wavelength is 1.3 m. If it's an open pipe then there are antinodes at the ends and the wavelength is 2 times the length, so length of the the pipe is about 1.3 m / 2 = .64 m, approx.. Had the pipe been closed at one end then there would be a node and one end and an antinode at the other and the wavelength of the fundamental would have therefore been 4 times the length; the length of the pipe would then have been 1.3 m / 4 = .32 m. **
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RESPONSE --> I got the formula L=v/2f out of the book. Is my answer incorrect?
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22:56:50 **** Univ phy 16.72 (10th edition 21.32): Crab nebula 1054 A.D.;, H gas, 4.568 * 10^14 Hz in lab, 4.586 from Crab streamers coming toward Earth. Velocity? Assuming const vel diameter? Ang diameter 5 arc minutes; how far is it?
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RESPONSE --> ok
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22:56:52 ** Since fR = fS ( 1 - v/c) we have v = (fR / fS - 1) * c = 3 * 10^8 m/s * (4.586 * 10^14 Hz) / (4.568 * 10^14 Hz) = 1.182 * 10^6 m/s, approx. In the 949 years since the explosion the radius of the nebula would therefore be about 949 years * 365 days / year * 24 hours / day * 3600 seconds / hour * 1.182 * 10^6 m/s = 3.5 * 10^16 meters, the diameter about 7 * 10^16 meters. 5 minutes of arc is 5/60 degrees or 5/60 * pi/180 radians = 1.4 * 10^-3 radians. The diameter is equal to the product of the distance and this angle so the distance is distance = diameter / angle = 7 * 10^16 m / (1.4 * 10^-3) = 2.4 * 10^19 m. Dividing by the distance light travels in a year we get the distance in light years, about 6500 light years. CHECK AGAINST INSTRUCTOR SOLUTION: ** There are about 10^5 seconds in a day, about 3 * 10^7 seconds in a year and about 3 * 10^10 seconds in 1000 years. It's been about 1000 years. So those streamers have had time to move about 1.177 * 10^6 m/s * 3 * 10^10 sec = 3 * 10^16 meters. That would be the distance of the closest streamers from the center of the nebula. The other side of the nebula would be an equal distance on the other side of the center. So the diameter would be about 6 * 10^16 meters. A light year is about 300,000 km/sec * 3 * 10^7 sec/year = 9 * 10^12 km = 9 * 10^15 meters. So the nebula is about 3 * 10^16 meters / (9 * 10^15 m / light yr) = 3 light years in diameter, approx. 5 seconds of arc is 5/60 of a degree or 5 / (60 * 360) = 1 / 4300 of the circumference of a full circle, approx. If 1/4300 of the circumference is 6 * 10^16 meters then the circumference is about 4300 times this distance or about 2.6 * 10^20 meters. The circumference is 1 / (2 pi) times the radius. We're at the center of this circle since it is from here than the angular diameter is observed, so the distance is about 1 / (2 pi) * 2.6 * 10^20 meters = 4 * 10^19 meters. This is about 4 * 10^19 meters / (9 * 10^15 meters / light year) = 4400 light years distant. Check my arithmetic. **
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RESPONSE --> ok
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22:56:54 **** query univ phy 16.66 (21.26 10th edition). 200 mHz refl from fetal heart wall moving toward sound; refl sound mixed with transmitted sound, 85 beats / sec. Speed of sound 1500 m/s. What is the speed of the fetal heart at the instant the measurement is made?
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RESPONSE --> ok
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22:56:56 . ** 200 MHz is 200 * 10^6 Hz = 2 * 10^8 Hz or 200,000,000 Hz. The frequency of the wave reflected from the heart will be greater, according to the Doppler shift. The number of beats is equal to the difference in the frequencies of the two sounds. So the frequency of the reflected sound is 200,000,085 Hz. The frequency of the sound as experienced by the heart (which is in effect a moving 'listener') is fL = (1 + vL / v) * fs = (1 + vHeart / v) * 2.00 MHz, where v is 1500 m/s. This sound is then 'bounced back', with the heart now in the role of the source emitting sounds at frequency fs = (1 + vHeart / v) * 2.00 MHz, the 'old' fL. The 'new' fL is fL = v / (v - vs) * fs = v / (v - vHeart) * (1 + vHeart / v) * 2.00 MHz. This fL is the 200,000,085 Hz frequency. So we have 200,000,085 Hz = 1500 m/s / (v - vHeart) * (1 + vHeart / v) * 2.00 MHz and v / (v - vHeart) * (1 + vHeart / v) = 200,000,085 Hz / (200,000,000 Hz) = 1.000000475. A slight rearrangement gives us (v + vHeart) / (v - vHeart) = 1.000000475 so that v + vHeart = 1.000000475 v - 1.000000475 vHeart and 2.000000475 vHeart = .000000475 v, with solution vHeart = .000000475 v / (2.000000475), very close to vHeart = .000000475 v / 2 = .000000475 * 1500 m/s / 2 = .00032 m/s, about .3 millimeters / sec. **
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RESPONSE --> ok
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