Download AP Physics B – Waves and Optics – FR 1 Answer Key SECTION A

AP Physics B – Waves and Optics – FR 1 Answer Key
SECTION A – Waves and Sound
#1 (1980-B4)
c) simple graph with 1.5x the frequency
d) Graphs are based on the Doppler equation. The graph given in the problem is for a moving observer.
Which is based on
f  f
vsnd  vobs  .
v snd
As the observer’s velocity increases, the frequency increases linearly
with it as is shown in the problem
The new graph is based on a source moving towards you.
f  f
v snd
. As can be seed from this
(v snd  vsource )
equation, as the source increases velocity, the frequency increases but when the source approaches the speed of
sound, the frequency approaches ∞ and becomes undefined so has a limit to it unlike in the first graph.
#2 (1995-B6)
a) The fundamental in a open–open pipe looks like this
this ½ wavelength fits in the length L, the total
and is ½ of a wavelength of the wave. Since
wavelength would have to be 2L.
b) Simply use v = f λ  v = 2Lfo
c) Harmonics are multiples of the fundamental, so the next frequency is 2fo
d) This is the same tuning fork so it is the same wavelength and waveform but the bottom is now closed so the wave
looks like this.
This is impossible for a standing
The tube we had initially, fit
wave in an open–closed tube, and
½ of a wavelength inside,
its also not the fundamental
and since its the same
anyway so we have to change the
wavelength wave, again ½ of
length to make it look like the
the wavelength of this wave
fundamental, Shown below. To do
would fit in length L and it
this, we make the length half of
would look like this.
what it used to be.
h = L/2
#3 (B2004-B3)
a) The shortest length makes the fundamental which looks like this
and is ¼ of the wavelength. This length
is known to be 0.25m. So L1 = ¼ λ … λ = 4L1 = 1m.
Note: This is a real experiment, and in the reality of the experiment it is known that the antinode of the wave
actually forms slightly above the top of the air column (you would not know this unless you actually performed this
experiment). For this reason, the above answer is technically not correct as the tube length is slightly less than 1/4
of the wavelength. The better way to answer this question is to use the two values they give you for each consecutive
harmonic. You are given the length of the first frequency (fundamental), and the length of the second frequency
(third harmonic). Based on the known shapes of these harmonics, the difference in lengths between these two
harmonics is equal to ½ the wavelength of the wave. Applying this 
∆L = ½ λ … 0.8–0.25 = ½ λ
λactual = 1.1 m.
Unfortunately the AP exam scored this question assuming you knew about the correction; though you received 3 out
of 4 points for using the solution initially presented. We teachers, the authors of this solution guide, feel this is a bit
much to ask for.
b) Using v = f λ with the actual λ … (340) = f (1.1)
… f = 312 Hz.
c) v = f λ … (1490) = (312) λwater … λwater = 4.8 m
d) Referring to the shapes of these harmonics is useful. The second length L3 was the 3rd harmonic. The next harmonic
(5th) will occur by adding another ½λ to the wave (based on how it looks you can see this). This will give a total
length of L2 + ½ λ = (0.8) + ½ (1.1) = 1.35 m
e) As temperature increases, the speed of sound in air increases, so the speed used in part (b) was too low.
Since f =vair / λ, that lower speed of sound yielded a frequency that was too low.
SECTION B – Physical Optics
#4 (1975-B4)
a) One slit
b) Two slits
wider
more narrow
#5 (1985-B5)
a) Simple application of the formula, m λ = d x / L … (1)(5x10–7) = (4x10–4)(x) / (2) … x = 2.5x10–3 m
b) i) n1 λ1 = n2 λ2 … (1)(5x10–7) = (1.3) λwater … λwater = 3.85 x10–7 m
ii) frequency does not change when changing mediums, same as air.
c = fair / λair … 3x108 = fair / 5x10–7 … fair = 6x1014 Hz = fwater
c)
Based on mλ = d x / L, since the λ is less, the x is less also which means the fringe spacing has decreased.
#6 (1980-B4)
a) Change double slits to diffraction grating
Notable features:
1) Maxima at same locations as before
(m λ = d sin θ ) would give same results
2) Maxima more narrow and well defined
3) Minimal intensity light in–between well
defined maxima’s
b) Spacing “d” of two slit arrangement doubled … m λ = d sin θ … 2x d  ½ the angle (for small angles)
First maxima occurs at half the distance away
as before.
Note: This diagram shows equal intensity
spots, in reality the intensity of the spots should
diminish slightly as moving away from center.
#7 (1991-B6)
a) m λB = d x / L … (1) (5.5x10–7) = d (1.2x10–2) / (0.85)
d = 3.9x10–5 m
b) m λa = d x / L … (1) (4.4x10–7) = (3.9x10–5) x / (0.85)
x = 9.6x10–3 m
#8 (1998-B7)
a) Diffraction grating: Since the screen distance is 1 m and the first order line is at 0.428 m, the angle is not small
and the small angle approximation cannot be used. Instead we find the angle with tan θ and use mλ = d sin θ.
First find d. d = 600 lines / mm = 1/600 mm / line = 0.00167 mm / line = 1.67x10–6 m / line.
tan θ = o/a … tan θ = 0.428 / 1 … θ = 23°
… Then, mλ = d sin θ
… (1) λ = (1.67x10–6) sin 23°
λ = 6.57x10–7 m = 657 nm
c) Referring to the calculation of d above .. d = 1/800 mm / line which is a smaller d value. Less d means sin θ must
increase so the angle is larger and the location of the line would be further out.
#9 (2004-B4)
Often with speakers, none of the approximation work and we simply have to work with the distances to find the path
difference, because the angle to the observer is not small and also the spacing of the speakers is also not small. In
this example, the spacing of the speakers is relatively small in comparison to the distance away L, so we can use
mλ = d sin θ. However the location of point Q is unclear so we will not assume that the small angle approximation
(x/L) would work.
a) Simple. v = f λ … 343 = 2500 f …
λ = 0.1372 m
b) Determine θ … mλ = d sin θ … (0.5) (0.1372) = (0.75) sin θ … θ = 5.25° (small enough to have used x/L)
Now find Y … tan θ = o / a … tan (5.25) = Y / 5 … Y = 0.459 m
c) Another minimum, ‘dark spot’ (not dark since its sound), could be found at the same distance Y above point P on
the opposite side. Or, still looking below P, you could use m = 1.5 and find the new value of Y.
d) i) Based on the formulas and analysis from point b, it can clearly be see that decreasing d, would make angle θ
increase, which would increase Y
ii) Since the speed of sound stays constant, increasing f, decreases the λ. Again from the formulas and analysis
in part b we see that less λ means less θ and decreases the location Y.
#10 (2005-B4)
a) Meterstick, Tape measure, Large Screen
b)
c)
d) Set up the laser to shine on the slide, and set the screen far away on the other side of the slide.
Measure the distance L from the slide to the screen with the tape measure. Use the ruler to measure the distance x
between adjacent maxima.
e) With the values obtained above, plug into m λ = d x / L, with m = 1 for the first spot and other variables as
defined above and the known λ of the laser used, solve for d. Assuming angle θ is small. It not, determine theta and
use m λ = d sin θ
#11 (B2005-B4)
This is basically the same as 2005B4 but with sound.
a) Meterstick, tape measure, sound level meter
b)
c) Set the speakers a fixed distance d apart, pointing perpendicular to the line along which d is measured.
Determine a line parallel to the speaker line and a distance L away. Use the sound meter to locate the maxima of the
interference pattern along this line. Record the locations of these, y values, maxima along the line.
d) With the values obtained above, plug into m λ = d x / L, with m = 1 for the first maxima and other variables as
defined above to determine the λ of the sound. Assuming angle θ is small. It not, determine theta and use
m λ = d sin θ. Then, with the λ plug into v = f λ with v as speed of sound to determine f.
e) Decreasing frequency results in increasing wavelength for constant velocity. Based on mλ = d x / L, larger
wavelength means a larger x value and thus the distance between successive maxima will increase.
#12 (2009-B6)
a) c = f λ … 3x108 = f (550x10–9) … f = 5.45x1014 Hz.
b)
m λ = d x1 / L … (0.5) (550x10–9) = (1.8x10–5) x / 2.2
m λ = d x2 / L … (1.5) (550x10–9) = (1.8x10–5) x / 2.2
x1 = 0.0336 m first dark spot
x2 = 0.101 m second dark spot
Distance between spots = .067 m
Alternatively you could find the location if the first bright spot from center and conclude the spacing of
consecutive bright and dark spots are equal so should all be spaced by this amount.