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

 

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:

 

 

 

 

 

 

 

 

 

 

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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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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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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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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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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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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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Confidence Rating:

 

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?

 

 

Your Solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

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:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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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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. 

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?

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Rating:

 

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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Confidence Rating:

 

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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