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HW |
Answers or clues to most homework
problems.
Yellow marks the current assignment. |
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#1 |
Q3
(a) #2
(b) positive |
Q7
(a) pi
(b) pi |
3
T = 0.5s, k = 79 N/m,
vm = 4.40 m/s,
Fm = 27.6 N |
12
x = 3 m, v = -49 m/s,
a = -270 m/s/s,
phase = 20 rad |
20
xm = 0.5m,
x = -0.251 m,
v = 3.06 m/s |
22
2.1 rad |
33
(b) 125 J
(c) 250 J
(d) Use one of the special points (an endpoint or equilibrium)
of SHM. |
36
xm =
mv / sqrt[k(M+m)] |
| S84 9.47 m/s/s |
S86 Solve for k in terms of m,g,h using
Newton's 2nd Law. |
| 55 froom = 0.35
Hz, felevator,up = 0.39 Hz, felevator,down
= 0 |
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| #2 |
Q5(a)
wave #4 |
2(a)
400 nm -> 7.5 x 10^14 Hz
700 nm -> 4.3 x 10^14 Hz |
S57
v = 200 cm/s |
S64
vt = 94 mm/s |
8
v = 2.0 cm/s; the phase constant in the wave equation is pi,
not zero; vt = -2.5 cm/s |
16
vm = 300 m/s |
S60
(a) P2 = 2*P1
(b) P2 = P1/4 |
26
phi = 1.45 rad
0.23 lambda |
33
v = 144 m/s
lambda = 60 cm
f = 241 Hz |
38
There are two harmonics that work. One is the 4th harmonic. |
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#3 |
Q14(a)
V3 > V2 = V1 |
3
the listener |
5
1900 km |
10
v = 6.0 m/s |
13
fmin,1 = 343 Hz and fmax,1
= 686 Hz. In the allowed range, there are 28 frequencies for
destructive interference (fmin,1 to fmin,28) |
16
r = 0.175 m |
22
39.5 dB |
28
The intensity at the microphone is 5.97x10^(-5) W/m^2 |
S75a
v = 170 m/s |
S76
(a) 77.6 Hz
(b) 77.0 Hz |
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#4 |
Q7
na > nb |
2
(a) 0.50 ms
(b) 503 s |
S82
B points in the positive x direction when E points in the positive
y direction (note that the light is moving in the negative z
direction).
Don't forget that all of our basic wave equation results apply
for light. We just need to use v = c = 3.00 x 10^(8) m/s. |
9
B = -y Bo cos (same phase factor
as E)
Bo = 6.7 x 10^(-9) T, f = 0.5 x 10^(15)
Hz |
45
For the observer to see the corner, a ray from the corner must
refract sideways when it hits the upper-left corner. n = 1.26. |
S69
diameter = 4.56 m |
53
nglass >=
1.22 |
57
You must block the rays that are not totally reflected.
63% gets covered. |
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#5 |
Q1
Decide first how the index of refraction should vary with height.
To do this, divide the air into horizontal layers and analyze
what happens between layers. Then use the fact that nair increases if the temperature decreases. |
1
40 cm |
6
1.5 m |
18
1.86 mm |
27
The first lens has an image that is real and inverted. This image
becomes the object for the second lens, and it is a virtual object
because it is on the "wrong" side. The final image
is virtual and has the same orientataion as the object. |
S60
Start by finding the image of the lens and ignoring the mirror.
That image becomes the object for the mirror. The mirror's image
becomes the new object for the lens, and it produces the final
image. The lens is used twice in this problem.
The final image is 0.6 m to the left of the lens, real, and not
inverted. |
27
The final image is in the same place as the object, virtual,
and inverted. |
35
For far objects, the retina must be at the focus. For
objects that are closer, it won't be. (a) 2.35 cm (b) radius
of curvature decreases. |
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#6 |
Q7a,b
decrease, decrease |
1
f = 5.09 x 10^(14) Hz |
5
22.0 degrees |
7
The wavelength in a medium that has index of refraction n, is
lambda/n. |
10
The smallest is 1550 nm. |
12
The phase difference for dark fringe m is (2m+1) pi. |
13
2.25 mm |
20
Determine the path length difference as a function of x. Answer
is:
8.75 times lambda |
21
"m" tells you the phase difference for a spot, however
the difference occurred. Answer = 6640 nm. |
27
The furthest is at x = 7.50 m. |
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#7 |
Q6
Assume the light diffracts through some structure in your
eye before reaching the retina.
(a) larger |
Q7
The larger a is, the smaller the diffraction central maximum.
(a) less |
2
0.00750 rad
0.118 mm |
4
a = 2.51 mm (that's big!) |
6
38 m |
10a
0.003143 rad |
9
Recall, I is proportional to E^2. First, break the slit into
N sections (Huygen wavelets). Each section contributes an electric
field wave in phase at the central max. Now, if you double the
slit size you'll have 2N sections. |
27,28
The spacing between 2-slit interference fringes is constant
to good approximation. So is the spacing between diffraction
minima.
Answer to 28: 3. |
31
Use your result from problem 28. |
32
d = 20.1 x 10^(-6) m
a = 5.04 x 10^(-6) m |
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#8 |
Answer Clues For Homework #8
Q4a
Sam |
Q10
(a) 3 > 2 = 1; (b) 2 > 1 = 3 |
S58
v = 0.75 c |
3
v = 0.99c |
6
v = 0.9999995 c |
11
87.4 m, 0.394 us |
25
1.2 us |
26
(a) 1.25 y; (b) 1.6 y; (c) 4.0 y |
37
18 suns/year |
S61
Use energy methods for both parts.
(a) 256 kV; (b) 0.746 c |
Some Small Hints For The Twin Paradox
| Q14 Use your answers to Q12. |
| Q16 Use your answers to Q15. |
| Q17 t = toutbound + tinbound,
so make sure to break the problem into two parts. |
| Q18 Use the definition of speed.
Since you already know the answer from Q7, the point here is
to show your work. |
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#9 |
5
Power is energy per unit time. You need to convert to photons
per unit time. |
8
8.6 x 10^5 m/s |
38
At what angle is the shift maximized?
2.65 x 10^(-15) m |
43
Back scattering means angle = 180 degrees.
(a) Find the change in wavelength first.
ans = 41.8 keV |
50
1.7 x 10^(-35) m |
64
(a) 15 keV |
74
Note we are given py
= pz = 0, so they
are given exactly. |
75
0.19 m |
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#10 |
Q2
Case (a) has the largest energy. |
Q4
This is a standing wave problem, right?
(a) Points of zero probability are places where the wave
function is always zero.
ans = 18 |
4
0.85 nm |
6
0.65 eV |
11
An electron in a well can only have certain allowed energies.
If it absorbs a photon, the photon's energy must be exactly that
necessary so that the electron ends up at one of these allowed
energies. This tells you what the allowed energies of the photon
are. Just convert photon energy to wavelength to complete the
problem. |
14
Look at Fig. 40-6.
(a) big decrease.
(b) small increase. (If you decided that the answer is "no
change", that's pretty close.) |
17
Use the hint provided in the text. (I'll work part of this in
class.) |
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