Assignment 15 Query

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course Phy 122

2/25 1 pm

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 evidence that the higher-pitched waves are longitudinal are found in the fact that in order to make the sound lessen by touching the rod is when you touch the very tip of the rod. The lower pitched waves are lessened by touching the sides of the end of the rod, not the very tip, which indicates that those are transverse waves.

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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):OK

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Self-critique Rating:OK

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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:

120 dB = 10 log(I/10^-12)

Intensity at 120 dB = 1 W/m^2

20 dB = 10 log (I/10^-12)

Intensity at 20 dB = 10^-10 W/m^2

Therefore the intensity at 120 dB is 10 billion times higher than the intesity at 20 dB.

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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