Please let me know if there are places where I am unclear, or
you think I am incorrect -- I do make arithmetic and typing mistakes
...
1. #8 on this jpg
2. The intensity equation is I = 0.5 * rho * v * omega2
* sm2 .
(a) So sm = SQRT [2 I / (rho * v * omega2) ]
= SQRT [2(10-1 W/m) / (1.21
kg/m * 343 m/s * (2*pi*1000 rad/s)2) ]
= 1.1e-11 m
(b) Since I increases by 1e-12, sm increases by 1e-6, sm
= 1.1e-5 m
(c) The answer to (a) is about 1/30 the size of an atom, and the answer to (b)
is about 4 million atoms.
3. Since I = 0.5 * rho * v * omega2 * sm2 , and the speed and density both depend only on the material the wave is travelling in, it will be larger frequencies that have smaller displacements amplitudes, for the same intensity.
4. (a) The wavelength = v/f = 343 m/s / 2000 Hz = 0.1715 m.
The path difference, 25.7 cm = 1.5 wavelengths, so the interference is fully
destructive.
(b) For fully destructive interference, the path difference must equal odd integral
multiples of (1/2) wavelengths:
path difference = (odd m) * lambda / 2 = (odd m) * v / 2f ...
f = (odd m) v / [2*path difference] = (odd m) * 343 m/s /
[2*0.257m] = 667 Hz * (odd m)
We know that for m = 3, f = 2000 Hz [part (a)], so we need m=5, f = 3.33
kHz.
(c) Change the separation of the two speakers: make the distance = 1 or 2 wavelengths
and you will have fully constructive interference: separation = 17.15 cm or
34.3 cm
5. We need to find how many extra wavelengths the sound
has to travel from the bottom speaker:
path length difference = distance from bottom speaker to hearer
- distance from top speaker to hearer
= SQRT[30^2 + 5.09^2] - 30 m = 0.429 m
We find the wavelenth from wavelength*frequency = speed:
wavelength = v / f = 343 m/s / 800 Hz = 0.429 m
The number of extra wavelengths is "exactly" one, so the two wave are in phase => fully constructive interference.
(b) Now to get fully destructive interference, the path difference
must equal either 1/2 or 3/2 the wavelength, depending on whether the speaker
is moved up or down.
Set SQRT [ 30^2 + x^2 ] - 30 = (3/2) 0.429 m or
30 - SQRT [ 30^2 + x^2 ] = (1/2) 0.429 m
x
= 6.245 m OR x
= 3.580 m
(c) If you double the frequency, you halve the wavelength. Thus, in (a) path difference is now equal to 2 wavelengths, and in (b), it is equal to 3 or 1 wavelengths, so both cases are fully constructive.
6. We never got to this question this quarter, but here are the answers.
(a) As the tube is pulled out, the sound is alternately loud and soft due to
alternating fully constructive and fully destructive interference.
(b) the change in the path difference is one wavelength => loud.
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8. #6 on this jpg
9. #7 on this jpg
This page maintained by Anne G. Young. Last modified 23-Apr-2005 .
