Assignment 15

course Phy 202

7/6 11:30p

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?

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Your solution:

Longitudinal waves move at a faster rate. The pitch of a wave can be found by the speed at which it reaches the ear. The longitudinal waves moved faster and thus had a higher pitch, while transverse waves moved slower and had a lower pitch.

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Given Solution:

`qSTUDENT RESPONSE: The logitudinal waves had a higher velocity.

`aThat 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):

&#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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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:

The equation we use is

dB= 10log(I/Io) Since we’re looking for intensity, our equation becomes

Io=threshold, which = 10^-12 watts/m^2

For 120 dB

log(I/Io)= dB/10 I/Io= 10^(120/10) I=Io*10^12

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

I= 1 watt/m^2

For 20 dB

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

I= 10^-10 watts/m^2

So 120 dB is (1 watt/m^2)/(10^-10 watts/m^2)= 10^10 times more intense than 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: `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:

First we find velocity of the sound waves

V= (331+0.60*21)

V= 343.6 m/s

Next we have to find the wavelength of the sound. The equation we have is frequency= velocity/wavelength. This can be manipulated to

Wavelength= velocity/frequency

Wavelength= (343.6 m/s)/(262 Hz)

Wavelength= 1.3 m

Because we have an open pipe, there will be antinodes at the ends and therefore the wavelength will be 2 times the length of the pipe.

So Length= (1.3m)/2

L= 0.65 m

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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. **

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Assignment 15

&#Very good work. Let me know if you have questions. &#