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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!