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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.
014. `Query 12
Question:
`qquery 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
Your solution:
Confidence Rating:
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. **
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Question:
`qquery General College Physics and Principles of
Physics: what is a decibel?
Your solution:
Confidence Rating:
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. **
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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
Your solution:
Confidence Rating:
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.
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.<
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Question:
`q **** gen phy
what are beats?
Your solution:
Confidence Rating:
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. **
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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?
Your solution:
Confidence Rating:
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).
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