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BROCK UNIVERSITY
Mid-term Test: March 2014
Course: PHYS 1P22/1P92
Examination date: 1 March 2014
Time of Examination: 11:00–12:30
Number of
Number of
Number of
Instructor:
pages: 7 (+ formula sheet)
students: 161
hours: 1.5
S. D’Agostino
A formula sheet is attached at the end of the test paper. No other aids are permitted except
for a non-programmable, non-graphing calculator.
Total number of marks: 25.
SOLUTIONS
1. [4 marks] The intensity of electromagnetic waves from the Sun is 1.4 kW/m2 just
above the Earth’s atmosphere. Eighty percent of this reaches the surface at noon on a
clear summer day. Suppose you model your back as a 30 cm × 50 cm rectangle. How
many joules of solar energy are incident on your back after 1.0 h?
Solution: The intensity of electromagnetic solar radiation that reaches the Earth’s
surface is
0.80 × 1.4 kW/m2 = 1.12 kW/m2
The area of your back is
0.3 m × 0.5 m = 0.15 m2
Thus, the total power of solar radiation incident on your back is
1.12 kW/m2 × 0.15 m2 = 0.168 kW = 0.168 kJ/s
There are 3600 s in 1 h, so the total amount of solar radiation received by your back
in 1 h is
0.168 kJ/s × 3600 s = 605 kJ
Thus, your back absorbs about 600 kJ of solar radiation in one hour.
Note the unspoken assumption in this problem: We assume that the sun’s rays are
incident on your back perpendicular to the plane of your back.
2. [4 marks] A guitar string with a linear density of 2.0 g/m is stretched between supports
that are 60 cm apart. The string is observed to form a standing wave with three
antinodes when driven at a frequency of 420 Hz.
(a) Determine the frequency of the fifth harmonic of the string.
(b) Determine the tension in the string.
Solution: (a) The wave with three antinodes has m = 3. Thus,
f3 = 3f1
and so the fundamental frequency of the string is
f1 =
420 Hz
f3
=
= 140 Hz
3
3
Thus, the frequency of the fifth harmonic is
f5 = 5f1 = 5 × 140 Hz = 700 Hz
(b) The speed v of a transverse wave on this string is related to the fundamental
frequency by
v
f1 =
2L
where L is the length of the string. Thus, the speed of a transverse wave on this string
is
v = 2Lf1 = 2(0.6 m)(140 Hz) = 168 m/s
The speed of the wave is related to the tension of the string and its linear density by
s
T
v=
µ
and so the tension in the string is
T = v 2 µ = (168 m/s)2 (0.0020 kg/m) = 56.45 N
3. [4 marks] A double-slit experiment is performed with light of wavelength 600 nm. The
bright interference fringes are spaced 1.8 mm apart on the viewing screen. Determine
the fringe spacing if the light is changed to a wavelength of 400 nm.
Solution: For the 600 nm light, we have
∆y1 =
λ1 L
d
whereas for the 400 nm light, we have
∆y2 =
λ2 L
d
Dividing the two previous equations (because L and d do not change), we have
∆y2
λ2 L/d
λ2
=
=
∆y1
λ1 L/d
λ1
and therefore
400 nm
∆y2
2
=
=
∆y1
600 nm
3
Thus, the fringe spacing when 400 nm light is used will be
2
2
∆y2 = ∆y1 = (1.8 mm) = 1.2 mm
3
3
4. [4 marks] A 100 g ball attached to a spring with spring constant 2.50 N/m oscillates
horizontally on a frictionless table. Its velocity is 20.0 cm/s when x = −5.00 cm.
(a) Determine the amplitude of oscillation.
(b) Determine the speed of the ball when x = 3.00 cm.
Solution: (a) First determine the total energy of the oscillator from the given information:
1
1
E = mv 2 + kx2
2
2
1
1
E = (0.100 kg)(0.200 m/s)2 + (2.50 N/m)(−0.05 m)2
2
2
E = 0.002 + 0.003125
E = 0.005125 J
When the displacement of the oscillator is equal to the amplitude, all of the oscillator’s
energy is potential:
1 2
kA = 0.005125 J
2
2(0.005125 J)
A2 =
k
2(0.005125
J)
A2 =
2.50 N/m
2
A = 0.0041 m2
A = 6.4 cm
(b) When the position is x = 3.00 cm = 0.0300 m,
1
1
E = mv 2 + kx2
2
2
2E = mv 2 + kx2
mv 2 = 2E − kx2
2E − kx2
v2 =
r m
2E − kx2
v=
m
s
v=
2(0.005125 J) − (2.50 N/m)(0.03 m)2
0.100 kg
v = 28.3 cm/s
5. [4 marks] An electric dipole is formed from ±1.0 nC point charges spaced 2.0 mm
apart. The dipole is centred at the origin and oriented along the y-axis.
(a) Determine the magnitude and direction of the electric field strength at the point
(x, y) = (10 mm, 0 mm).
(b) Determine the electric potential at the point (x, y) = (10 mm, 0 mm).
Solution: (a) Assume that the positive charge is at (0, 1.0 mm) and the negative
charge is at (0, −1.0 mm). Of course, you could also make the opposite assumption,
because the statement of the problem doesn’t specify this; if you make the opposite
assumption, the electric field in the result will have the same magnitude but opposite
direction.
Draw a diagram!
In your diagram, you’ll notice a key angle,
1
−1
= 5.71◦
θ = tan
10
The distance r from each of the charges to the point (10 mm, 0 mm) at which we wish
to calculate the field satisfies
2
2
r2 = 1 × 10−3 + 10 × 10−3 = 1.01 × 10−4 m2
The electric field created by the positive charge at the point (10 mm, 0 mm) is
KQ
KQ
KQ
~1 =
E
cos θ, − 2 sin θ = 2 (cos θ, − sin θ)
2
r
r
r
The electric field created by the negative charge at the point (10 mm, 0 mm) is
KQ
KQ
KQ
~
E2 = − 2 cos θ, − 2 sin θ = 2 (− cos θ, − sin θ)
r
r
r
~1 + E
~ 2:
By the principle of superposition, the total electric field is the vector sum E
~ =E
~1 + E
~2
E
~ = KQ (cos θ, − sin θ) + KQ (− cos θ, − sin θ)
E
r2
r2
~ = KQ (cos θ − cos θ, − sin θ − sin θ)
E
r2
~ = KQ (0, −2 sin θ)
E
r2
The magnitude of the electric field is
KQ
(2 sin θ)
r2
(8.99 × 109 ) (1 × 10−9 )
E=
(0.199)
1.01 × 10−4
E = 18 kV/m
E=
The electric field is in the −y direction.
Remember that if you placed the dipole in the opposite direction to the one chosen
here then the magnitude of the electric field will be the same as calculated here, but
the direction of the electric field will be in the +y-direction.
(b) Use the superposition principle; that is, determine the potential at the given point
due to each point charge separately, and then add the results to obtain the total
potential.
The distance from each point charge to the given point can be determined using
Pythagoras’s theorem:
√
√
r = 12 + 102 = 101 = 10.05 mm = 0.01005 m
The potential at the given point due to the positive point charge is
Kq1
r
(9 × 109 )(1.0 × 10−9 )
V1 =
0.01005
V1 = 896 V
V1 =
The potential at the given point due to the negative point charge is
Kq2
r
(9 × 109 )(−1.0 × 10−9 )
V2 =
0.01005
V2 = −896 V
V2 =
The total electric potential at the given point, due to both point charges, is
V = V1 + V2 = 896 − 896 = 0 V
Compare the solution method here to the solution in Part (a). Notice the key difference
that arises because electric potential is a scalar, whereas the electric field is a vector.
Thus, the electric potential has no components, and the solution in Part (b) is easier
than the solution in Part (a). This foreshadows something that all you physics majors
will be studying in detail in your third-year electricity and magnetism courses: If
you need to calculate the electric field created by some charge configuration, often
the easiest way is to first calculate the electric potential, and then differentiate the
potential (determine the negative gradient of the potential, if we are being technical)
to determine the electric field.
6. [5 marks] Circle the best response in each case.
(a) Light of wavelength 500 nm is incident on a diffraction grating, producing an
interference pattern on a screen. If the wavelength increases to 600 nm,
i.
ii.
iii.
iv.
v.
the width of individual bright lines in the interference pattern increases.
the width of individual bright lines in the interference pattern decreases.
the spacing between bright lines in the interference pattern increases.
the spacing between bright lines in the interference pattern decreases.
there is no change in the interference pattern, because the spacing between
the lines in the grating does not change.
Answer: iii
(b) If the length of a tube increases, the fundamental frequency of a standing sound
wave in the tube
i. increases.
ii. decreases.
iii. does not change, because the fundamental frequency depends on the speed of
sound.
iv. does not change, because the fundamental frequency depends on the value of
the mode number.
v. does not change, because the fundamental frequency depends only on the
type of tube, not its length.
Answer: ii
(c) A simple harmonic oscillator with no damping has mass m and stiffness constant
k. The oscillator is displaced from equilibrium by an amount A and then released.
The total energy of the oscillator depends on
i.
ii.
iii.
iv.
v.
the
the
the
the
the
frequency of the oscillation.
time since the oscillation began.
ratio of the stiffness constant and the mass.
square root of the ratio of the stiffness constant and the mass.
amplitude of the oscillation.
Answer: v
This question is poorly worded, and I ended up giving a point to everyone for this
question, no matter what they answered. The point I was trying to make is that
the formula for the total energy of a simple harmonic oscillator does not depend
on the frequency of oscillation. However, I didn’t succeed in making the point
very well.
Here’s what I was thinking: If you change the stiffness constant of the spring, and
the mass of the oscillator, but do not change their ratio, then the total energy of
the oscillator does not change if you leave everything else the same. This means
that (i), (iii), and (iv) are ruled out. There is no damping present, so this rules
out (ii). Thus, by process of elimination, (v) is the correct response. And indeed,
the total energy of the oscillator is equal to (1/2)kA2 .
(d) Electric field lines
i. are close together where the electric field is strong, and far apart where the
electric field is weak.
ii. point away from positive charges and towards negative charges.
iii. are such that electric field vectors are tangent to them.
iv. never cross where the electric field is nonzero.
v. [All of the above are true].
Answer: v
(e) Electric equipotential surfaces
i. are close together where the electric field is strong, and far apart where the
electric field is weak.
ii. point away from positive charges and towards negative charges.
iii. are such that electric field vectors are tangent to them.
iv. are such that electric potential vectors are tangent to them.
v. [All of the above are true].
Answer: i
Remember that the electric potential is a scalar, not a vector!