Assignment 15

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course PHY 232

If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://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.

015. `Query 13

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Question: `qquery 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?

Your Solution:

In case of longitudinal waves, the sound seems to reach the ear earlier.

Confidence Rating:

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

I was not able to explain the concept in this one clearly. It seems that I haven’t really grasped the concept or maybe I’m having trouble answering in an appropriate way.

@&

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

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Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

• 2 indicates that you believe you addressed most of the discrepancies between the given solution and your solution but are unsure of some aspects of the situation; you would at this point consider including a question or a statement of what you're not sure you understand

• 1 indicates that you believe you understand the overall idea of the solution but have not been able to address the specifics of the discrepancies between your solution and the given solution; in this case you would normally include a question or a statement of what you're not sure you understand

• 0 indicates that you don't understand the given solution, and/or can't make a reasonable judgement about whether or not your solution is correct; in this case you would be expected to address the given solution phrase-by-phrase and state what you do and do not understand about each phrase)

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Given Solution: `qSTUDENT 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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Question: `qquery 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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Your solution:

dB = 10 log(I / I_threshold) for I.

log(I / I_threshold) = dB / 10

I / I_threshold = 10^(120 / 10) = 12and

I = I_threshold * 10^12.

I = 10^-12 watts / m^2 * 10^12 = 1 watt / m^2.

= 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).

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.

Confidence Rating:

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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Given Solution:

`aThe 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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Question: Openstax: Dolphins make sounds in air and water. What is the ratio of the wavelength of a sound in air to its wavelength in seawater? Assume air temperature is 20.0ºC .

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Your Solution:

Since the speed of sound in water is roughly 5 times that in air, the same number of peaks will be distributed over about 5 times the distance in water as opposed to in air. So the wavelength in water will be about 5 times that in air.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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Given Solution: Using propagation speeds 340 m/s and 1600 m/s for sound in water and air, respectively:

The frequency of a particular sound is determined by the frequency at which the vocal cords vibrate, and is the same in water as in air. Since the speed of sound in water is roughly 5 times that in air, the same number of peaks will be distributed over about 5 times the distance in water as opposed to in air. So the wavelength in water will be about 5 times that in air.

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Question: Openstax: The warning tag on a lawn mower states that it produces noise at a level of 91.0 dB. What is this in watts per meter squared?

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Your Solution:

dB = 10 log( I / I_0)

log( I / I_0) = dB / 10

given dB = 91

Hence

log( I / I_0) = dB / 10

I / I_0 = 10^9.1 = 1.3 * 10^9

I = 1.3 * 10^9 * I_0 = 1.3 * 10^9 * (10^-12 watts / m^2)

Hence,

1.3 * 10^-3 watts / m^2 is the answer.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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Given Solution:

dB = 10 log( I / I_0), so

log( I / I_0) = dB / 10.

In this case dB = 91 so

log(I / I_0) = 91/10 = 9.1.

It follows that

I / I_0 = 10^9.1 = 1.3 * 10^9, approx..

so that

I = 1.3 * 10^9 * I_0 = 1.3 * 10^9 * (10^-12 watts / m^2) = 1.3 * 10^-3 watts / m^2,

or about .0013 watts / m^2.

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Question: Openstax: (a) What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound?

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Your Solution:

Intensity of a 90dB sound is

I = 10^9 * (10)^(-12)

For twice the intensity,

dB = 10 log( I / I_0)

= 10 log(2 * 10^9 * I_0 / I_0)

= 10 log( 2 * 10^9) = 10 * 9.3 = 93.0

For 1/5 intensity,

dB = 10 log( I / I_0)

= 10 log(0.2 * 10^9 * I_0 / I_0)

= 10 log( 0.2 * 10^9)

= 10 * 8.3

= 83.0

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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Given Solution: A 90-dB sound has intensity I = 10^9 * I_0, where I_0 is the hearing threshold intensity 10^-12 watts / m^2.

For a sound of twice that intensity I = 2 * 10^9 * I_0, the dB level is

dB = 10 log( I / I_0) = 10 log(2 * 10^9 * I_0 / I_0) = 10 log( 2 * 10^9) = 10 * 9.3 = 93.

For a sound 1/5 that intensity

dB = 10 log( I / I_0) = 10 log(0.2 * 10^9 * I_0 / I_0) = 10 log( 0.2 * 10^9) = 10 * 8.3 = 83.

Note that the first sound is 10 times as intense as the second (double a number is 10 times as great as 1/5 that number), and the dB difference between these sounds is 10, adding validation to the principle that a sound 10 times as intense as another exceeds the former to 10 dB.

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Question: Openstax: (a) What is the fundamental frequency of a 0.672-m-long tube, open at both ends, on a day when the speed of sound is 344 m/s? (b) What is the frequency of its second harmonic?

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Your Solution:

The fundamental mode of vibration has anti-nodes at both ends.

Length:

L(lambda) = 2 * .672 m = 1.34 m

Frequency = 344/1.34

Wavelength of second harmonic:

L_2(second fundamental) = .67 m

Frequency = 344/1.34

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*:: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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Given Solution:

If a tube is open at both ends, its fundamental mode of vibration has an antinode at both ends, with no additional antinodes. The configuration of nodes and antinodes is thus ANA, which corresponds to one-half wavelength. Thus the length of the tube is one-half wavelength, and the wavelength in this case is

lamba_fundamental = 2 * .672 m = 1.34 m, approx.. (the positions of the antinodes are not exactly at the ends of the tube, so a four-significant-figure result would imply more precision than can be expected).

Every harmonic has antinodes at the ends. The second harmonic has an antinode in the middle as well, with a node (as always) between two adjacent antinodes. So the configuration of this harmonic is ANANA, corresponding to a full wavelength. The wavelength of the second harmonic is thus

lambda_2d_harmonic = .67 m, approx..

The frequencies are easily calculated from the wavelengths and the 344 m/s propagation velocity.

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Question: Openstax: (a) Ear trumpets were never very common, but they did aid people with hearing losses by gathering sound over a large area and concentrating it on the smaller area of the eardrum. What decibel increase does an ear trumpet produce if its sound gathering area is 900 cm^2 and the area of the eardrum is 0.500 cm^2 , but the trumpet only has an efficiency of 5.00% in transmitting the sound to the eardrum? (b) Comment on the usefulness of the decibel increase found in part (a).

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Your Solution:

Sound gathering area of ear trumpet = 900 cm^2

Area of eardrum = 0.500 cm^2

Efficiency = 5%

The sound is gathered from an area which is 900 cm^2 / (0.5 cm^2) = 1800 multiplied by the area.

This would result in 1800 times the intensity.

Since the efficiency is only 5%, the actual intensity would be 5% of 1800 = 90

Since the intensity is increased, it could help in boosting the sound where required, also to people who have hearing loss.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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Given Solution: The sound is gathered from an area which is 900 cm^2 / (0.5 cm^2) = 1800 times the area, which if transmitted without loss to the eardrum would result in 1800 times the intensity.

However transmission is only 5% efficient, so the increase in intensity is by a factor of only 5% of 1800 or 90.

Every 10 times the intensity results in a 10 dB increase. 90 is almost 10 times 20, so the increase in intensity would be nearly 20 dB.

This could be quite useful for someone with hearing loss, though modern hearing aids do much better.

Note: A more precise estimate that 20 dB would be useless in this context, since eardrums vary significantly in size (0.500 cm^2 implies 3-significant-figure accuracy, which is absurd in this context). However more precise calculations are possible. In this case, dB = 10 log(90) would yield 19.5 dB rather than 20, and the .5 dB difference would be insignificant.

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Question: Openstax: What frequencies will a 1.80-m-long tube produce in the audible range at 20.0ºC if: (a) The tube is closed at one end? (b) It is open at both ends?

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Your Solution:

The node and antinodes will be placed like

Node-Antinode,

Node-Antinode-Node-Antinode…and so on

with the length of the tube being 1/4, 3/4, 5/4, ... times the wavelength of the sound.

Possible wavelengths will therefore be 4, 4/3, 4/5, 4/7, ... . times the length of the tube

Longest possible wavelength: 4 * 1.8 m = 7.2 m

Frequency produced:

v/7.2 = 50 Hz approx..

4/3 * 1.8 m = 2.4 m, f =v/2.4

4/5 * 1.8 m = 1.45 m, f=v/1.5

4/7 * 1.8 m = 1.03 m f=v/1.03

If it is open at both ends:

Antinode-Node-Antinode,

Antinode-Node-Antinode-Node-Antinode…and so on

and the length of the tube will be equal to 2, 1, 2/3, 1/2, ... times the wavelength of the sound (these numbers come from the fact that the respective node-antinode configurations comprise 2/4, 4/4, 6/4, 8/4, ... of the length of the tube).

Doing the same procedure we can get the frequencies.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 2

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Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

My answer was not that detailed.

@&

&#This also requires a self-critique.

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Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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Given Solution:

We reason out our results by considering possible node-antinode configurations.

If the tube is closed at one end then the possible node-antinode configurations are

NA, NANA, NANANA, NANANANA, ...,

with the length of the tube being 1/4, 3/4, 5/4, ... times the wavelength of the sound.

Possible wavelengths will therefore be 4, 4/3, 4/5, 4/7, ... . times the length of the tube.

Assuming propagation speed 340 m/s and audible range 30 Hz - 15 000 Hz, we easily see that the longest possible wavelength, which is 4 * 1.8 m = 7.2 m, will produce a sound around 50 Hz, clearly in the audible range.

The next few wavelengths are 4/3 * 1.8 m = 2.4 m, 4/5 * 1.8 m = 1.45 m, 4/7 * 1.8 m = 1.03 m (approx.), etc.

A 15 000 Hz sound would have wavelength 340 m/s / (15 000 s^-1) = 2.3 * 10^-2 m, or .023 m. At this wavelength the tube spans a little more than 78 wavelengths.

78 = 312 / 4, so if the tube is 311 / 4 times the wavelength the wavelength will be at least .023 m and have a frequency less than about 15 000 Hz. If the tube length is 313/4 times the wavelength the frequency will exceed 15 000 Hz.

Thus if, and only if, tube length is equal to 1/4, 3/4, 5/4, ..., 311/4 times the wavelength, the wavelength will produce audible sound.

Audible wavelengths will therefore be 4 * 1.8 m, 4/3 * 1.8 m, 4/5 * 1.8 m, ..., 4/311 * 1.8 meters. The corresponding frequencies can be obtained by dividing the speed of sound by each wavelength.

For example, wavelength 4/311 * 1.8 meters implies frequency 340 m/s / (4/311 * 1.8 meters) = 14700 Hz, approx..

If the tube is open at both ends then the node-antinode configurations are

ANA, ANANA, ANANANA, ...

From the above reasoning we see that the tube will produce audible sounds up to the harmonic where tube length is 312/4 times the wavelength of the sound.

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Question: `qquery 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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Your solution:

Wavelength = velocity/frequency

v = (331 + 0.60 * Temp.) m/s

= (331 + 0.60 * 21) m/s

= 343.6 m/s

Hence,

Wavelength = 343.6/262 = 1.33 meters

For length of the pipe:

1/4 * 1.33 meters = .33 meters, when it is between a node and an adjacent anti-node.

For total of two node and anti-node distances, the total length of pipe would be : 2*0.33 = 0.66 meters

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

.............................................

Given Solution:

`aGOOD 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) = 1.33 meters, approx..

The pipe is open, so it has antinodes at both ends.

• The fundamental frequency occurs when there is a single node between these antinodes. So the length of the pipe corresponds to two node-antinode distances.

• Between a node and an adjacent antinode the distance is 1/4 wavelength. In this case this distance is 1/4 * 1.33 meters = .33 meters, approx..

• The two node-antinode distances between the ends of the pipe therefore correspond to a distance of 2 * .33 meters = .66 meters.

We conclude that the pipe is .64 meters long.

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 1.33 m / 4 = .33 m.

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Question: `q**** Univ phy 16.79 11th edition 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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Your solution:

fR = fS ( 1 - v/c)

Getting value of v:

v = (fR / fS - 1) * c =

3 * 10^8 m/s * ( 4.586 * 10^14 Hz / (4.568 * 10^14 Hz) - 1) =

1.182 * 10^6 m/s, approx

949 years * 365 days / year * 24 hours / day * 3600 seconds / hour * 1.182 * 10^6 m/s

= 3.5 * 10^16 meters = radius of nebula

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Response: 2

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• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

I did not know how to proceed after the middle of the solution.

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Given Solution:

`a** 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) = 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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Question: `q **** query univ phy 16.74 / 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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Your solution:

2mHz is 2*10^8 Hz

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.

fL = (1 + vL / v) * fs

= (1 + vHeart / v)

fL = v / (v - vs) * fs = v / (v - vHeart) * (1 + vHeart / v) * 2.00 MHz

(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

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Given Solution:

`a. ** 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. **

STUDENT COMMENT

My final answer was twice the answer in the given solution. I thought that I used the

Doppler effect equation correctly; however, I may have solved for the unknown incorrectly.

INSTRUCTOR RESPONSE

The equations tell you the frequency that would be perceived by a hypothetical detector on the heart.

Suppose that each time the detector records a 'peak', it sends out a pulse. The pulses are sent out at the frequency of the detected wave. The source of these pulses is the detector, which is moving toward the 'listener', and as a result they are detected at an even higher frequency.

Thus the doubled number of beats.

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