Query 15

course Phys 202

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:

The high pitch waves were easily quieted when touching the end of the rod. The frequency interpreted by the ear is what classifies pitch. Intensity is proportional to the Amplitude^2. The L waves had the higher velocity.

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):

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Self-critique rating #$&*

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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:

Beta= 10 (log) (I/ I 0)

I 0= 1x10^-12

Solve for I:

10^(120/10) (1x10^-12)= 1 W/m^2

And

10^(20/10)(1x10^-12)= 1x10^-10 W/m^2

Ratio: 1W/m^2/ 1x10^-12 W/m^2= 1x10^10

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):

OK

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Self-critique rating #$&* 3

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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:

Open Pipe= v/2L

V= 331 +0.6*21= 343.6 m/s

Lambda= vel/freq= 343.6/262= 1.31 meters

According to the open pipe formula= length is twice as large..so 1.31/2=0.65 meters

confidence rating #$&*

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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) = 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. **

I've revised the above solution, which came up with the right answer (the length of the pipe is about .64 meters, the node-antinode distance is about .32 meters and the . When editing I didn't notice that there are a couple of lines that aren't completely coherent. And it was possible to insert a bit more explanation into the given 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 1.3 m / 4 = .33 m.

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Self-critique (if necessary):

Where is the 0.33 coming from??? When I divide 343.6/262, I get 1.31meters…

The revised solution, included above, should clarify this.

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Self-critique rating #$&*3

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&#Good responses. See my notes and let me know if you have questions. &#