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#### Transcript of Reading Assignment 9

2) Wave fronts represent the crests of the waves. What is the phase difference between two adjacent wave fronts?
c) 2pi
We know that the distance between crests of two waves is one definition of a wavelength and we also know that one wavelength equals to 2pi.
3) Look at Example 15-4, Fig 15-18. What would the new path difference, Δd, be for the point P of the observer if it was located at (x,y)=(3.0m, 1.0m)? Hint: DRAW a new figure to help visualize the geometry.
c) 0.6m
Phase difference is simply d2-d1 (
larger distance subtracted by the smaller one
). We know from the diagram, d1 originated from (0,1) and d2 originated from (0,-1). If the new point P is (3,1), then
d1 would be 3m
. D2 would have y-component of 1-(-1) = 2 and x component of 3m. Using phythagorean theorem, we will get
d2 equal to square root of 13
.
d2-d1 = 0.6m
4) For this question, I will instead discuss Figure 15-14 from the textbook
As mentioned, red regions represent large positive amplitude. This also means that they represent the crest (top part) of the wave. Meanwhile, green regions represent large negative amplitude. This also means that they represent the trough (bottom part) of the wave. Both red and green correspond to maximum intensity. Note that intensity is correlated to amplitude squared. The black area contains a mix of crest and trough, meaning that destructive interference occurs in that region.
5) A tuning fork and a piano emit f1 = 523 Hz and f2 = 527 Hz, respectively. What do you hear?
f) A sound at 525 Hz with 4 beats per second
We know that to get the
beat frequency
, we simply
subtract the two frequencies
, resulting in 4 Hz, the amplitude of the resultant wave would be modulated at 4 Hz --> 4 beats per second. As for the
sound frequency
, we take the
average of the two
, resulting in 525 Hz.
Choose ALL the correct statements about beats
a)
The slow modulation causes the soft-loud-soft "beat" pattern we hear
b)
The slow modulation of frequency is described by a sinusoid
c)
When the two frequencies are nearly equal, we hear only the average oscillation frequency
f)
Typically favg >> fbeat

a) Looking at figure 15-21, we see that the amplitude modulation correspond with a beating. Also, the reason why we observe a soft-loud-soft beat pattern is because as seen on the graph, the
amplitude increases then decreases within a modulation period
. Amplitude corresponds with "loudness", thus, we see the "beat" getting louder and softer and so on.
b) Again, we can look at the same graph to determine that modulation of frequency is modeled by a sinusoidal graph.
c) When the two frequencies are nearly equal,
beating frequency is very minimum,
resulting in us almost hearing just the average oscillation frequency. In other words, we won't be hearing f1 and f2 independently
f) favg = (f1+f2)/2 fbeat = (f2-f1)
putting any numbers into these formula, we will always get favg >> fbeat. Try it!
7) Looking at Checkpoint C-28-2 in the textbook (pg 766), how many times will the centre of the output go dark if you move one of the mirrors by 1.0mm? Choose the closest answer.
d) More than 3000 times
1.0 mm = 1,000,000 nm. We know that each distance d that the mirror moves, the
change in path difference distance is twice that value
, so path difference is 2,000,000 nm. We also know that the centre of the output goes dark (
complete destructive interference) every 0.5 lamda
. We are given that lamda is 632 nm, so 0.5 lamda is 316 nm. Therefore, we have 2,000,000nm/316nm = more than 300 times
Discussions

1) How far should speaker 2 be away from speaker 1 in order to completely cancel out the sound at the microphone (i.e., cause destructive interference)?
d) eitherλlamda/2 or 3lamda/2
The reason why d) is correct is because both will cause the plane waves to be
perfectly out of phase
, thus resulting in
destructive interference
. You should also note that the path differences are
half-integer multiples
of the wavelengths.

b) and e) are technically correct, but on their own, they imply that there is no other possibility

a) and c) are incorrect as they will cause the waves to remain perfectly in phase, resuting in constructive interference where their amplitudes add. Note that the path differences are integer multiples of the wavelenghts
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