Query 14

#$&*

course Phy 122

2/23 2237hrs.

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

When the side was struck, I heard the higher pitch, just not as strong as the tapping the end of the rod. After striking the side, and then touching it, I could still hear the higher pitch faintly but if the tip was touched, the high pitch goes away faster. So even after cranking my hearing aids up all the way and putting their blue-tooth microphone right next to my computer, I witnessed the same results. After replaying it a few times, I concluded that the higher pitch waves were traveling longitudinal and were damped a lot quicker by touching the ends. By touching the sides, the tranverse waves (lower pitch) were damped much quicker. Then my dog started acting weird.

confidence rating #$&*:

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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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Self-critique (if necessary):

Nailed it on the first part of the given solution. Curious, I have blast over-pressurization in both ears so the hairs behind my ear-drum is permanently curved towards the ear drum resulting in hyper accusative hearing causing sound and pitch comprehension issues. Thus, I where noise dampening hearing aids to block out back ground noise. With my aids I believe that I heard was justified by my eyes as the professor touched the rod. Without my aids I could really tell if the sounds were my speakers fuzz or the actual pitch. But my dog acted the same way for each trial.

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Self-critique Rating:3

@&

I have hearing issues as well, but they're age-related. I'm losing the high pitches. I hear the high pitch fine, but my students clearly find it a lot more prominent than I do.

I'll have to check this out with our dog.

We can produce higher pitches as well by supporting the rod at appropriate points. I can't hear them, but my students can. It would be interesting to try to get to the harmonics beyond the human range but within the dog's.

*@

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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:

120dB -20dB = 100dB

100/10=10

120 / 20 = 10 so 10^10 higher intensity for 120 dB than 20dB

@&

Good. For full clarity you would want to say 10^10 times the intensity.

*@

confidence rating #$&*:

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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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Self-critique (if necessary): Not sure about the top explanation, I will look at that a little further, but the bottom part is where I began, and that looks correct.

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Self-critique Rating:2

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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:

As a rule of thumb, as an EOD diver, our handheld sonar searched use a ½ the distance, 4 times the intensity rule. I know that sound travels 4 to 5 times faster in water and the echo location of an atlantic bottle nose dolphin is 5 times that of my sonar so I would say 4 to 5 times faster in water. Not sure what the air temp would imply.

confidence rating #$&*:

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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.

Your Self-Critique: don’t know how I got this one.

@&

Experience, obviously.

*@

Your Self-Critique Rating:2

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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:

91/10 = 9.1

10^9.1 = 1.26

(10^9.1)(10^-12 W/m^2) = 10^-2.9 W/m^2

confidence rating #$&*:

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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.

Your Self-Critique:should have rounded 9.1 to 9 and 1.26 to 1.3.

Your Self-Critique Rating:3

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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:

90/10 = 9, so I = 10^9

2*10^9 = 10 log (2000000000) = 9.3*10 = 93dB

1/5 = .2

.2 * 10^9 = 10 log (200000000) = 8.3*10 = 83dB

confidence rating #$&*:

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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.

Your Self-Critique: NA

Your Self-Critique Rating: 2

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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:

Lambda = 2l = 2(.672m) = 1.34 m/2 = .67m

Lambda = 344m/s / x Hz = .67m so, 344/.67 = 513.43Hz

confidence rating #$&*:

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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.

Your Self-Critique: I need to readdress understanding the node/antinode relationship.

@&

It's actually pretty simple. A node is followed by an antinode which is followed by a node. So if you know whether each end is a node or an antinode, you can write down the possible sequences of nodes and antinodes.

*@

Your Self-Critique Rating:3

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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:

900/.5 = 1800

.05 * 1800 = 90

I would imagine that increasing the dB would help the hearing impaired who suffer from the lack of normal elasticity in the ear drum, (It doesn’t vibrate like it should). By intensifying the sound, and increase in dB would cause a damaged ear drum to bounce more resulting in a perceived louder noise level.

confidence rating #$&*:

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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.

Your Self-Critique: NA

Your Self-Critique Rating:NA

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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: going back to the book on this one. I couldn’t even guess.

confidence rating #$&*:

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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, ...

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).

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.

Your Self-Critique: This is a tough one, I will have to print this out and try it a few times.

Your Self-Critique Rating:1

@&

A closed end has an node, and open end has a node.

So the possible configurations are

NA

NANA

NANANA

etc..

each node-antinode distance is 1/4 wavelengths.

The rest follows from these premises.

*@

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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: going back to the book on this one. I couldn’t even guess.

confidence rating #$&*:

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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, ...

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).

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.

Your Self-Critique: This is a tough one, I will have to print this out and try it a few times.

Your Self-Critique Rating:1

@&

A closed end has an node, and open end has a node.

So the possible configurations are

NA

NANA

NANANA

etc..

each node-antinode distance is 1/4 wavelengths.

The rest follows from these premises.

*@

#*&!

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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: going back to the book on this one. I couldn’t even guess.

confidence rating #$&*:

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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, ...

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).

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.

Your Self-Critique: This is a tough one, I will have to print this out and try it a few times.

Your Self-Critique Rating:1

@&

A closed end has an node, and open end has a node.

So the possible configurations are

NA

NANA

NANANA

etc..

each node-antinode distance is 1/4 wavelengths.

The rest follows from these premises.

*@

#*&!#*&!

&#Your work looks good. See my notes. Let me know if you have any questions. &#