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Position: # (your seat #)………………………………………….. Your Name: Last, First…………………………………………… Name of Workshop Instructor: Last, First……………………….. Exam: # (#1)……………………………………………………… Date: February 10 th, 2005………………………………………… Q1. The equation of a wave is y = 5.6 sin [0.5 (x-56t)/3] Here, both x and y are in centimeters. Also, t is in seconds. Find: a) the amplitude of the wave; b) the wavelength of the wave; c) the frequency of the wave; d) the speed of the wave; e) the period of the wave. Answer: The full form of the wave is: y = A sin [(2πx/λ) – (2πt/T) The expression written above can be rewritten: y = 5.6 sin [0.5 (x- 56t)/3]=5.6 sin (0.5 x/3-28 t/3) a) Therefore, the amplitude of the wave is A=5.6 cm b) From 0.5 x/3=2πx/λ The wavelength is λ=12 cm c) From 28 t/3=2πt/T, T=6/28 s, f=1/T=28/6 s -1 =4.6 s -1 d) The speed of the wave is v=λ/T=(12/6)×28=56 cm/s e) From 28 t/3=2πt/T, the period is T=6/28=0.21 s

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Q2. Write an equation for a traveling wave whose amplitude is 25 cm, wavelength is 45 cm, wave-speed is 10 meters per second, and which is traveling in the positive x direction. Answer: The full form of the wave is: y = A sin [(2πx/λ) – (2πt/T)] Where A=25 cm, λ=45 cm, v=10 m/s=1000 cm/s If v=λf= λ/T, then T=λ/v=45/1000=0.045 s Therefore, the full form of the wave is: y= 25 sin [(2πx/45)-(2πt/0.045)] Q3. An oscillating elastic system undergoes a simple harmonic motion. The elastic force corresponding to a displacement x = 25 cm from the equilibrium position is F = 76 N. Find: a) the elastic constant k; b) the elastic force that corresponds to a displacement x = 58 cm from the equilibrium position; c) the displacement x that corresponds to an elastic force of 10 N. Answer: F = kx a) Then, the elastic constant is k=F/x=76/25 N/cm=3.04 N/cm b) The elastic force is F=kx= 3.04×58=176.32 N c) The displacement x=F/k=10/3.04 cm=3.29 cm

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Q4. The unit for the product of wavelength and frequency is: a) ms; b) Hz; c) s/m; d) m/s. Pick the best choice for the answer, and explain your reasoning. Answer: [ ]×[f]=m × s -1 = m/s. Therefore, the correct answer is d). Q5. The lower the frequency of an electromagnetic wave in vacuum: a) The smaller is its wave-speed; b) the larger is its wave- speed; c) the shorter is its wavelength; d) the longer is its wavelength; e) the longer is its period; f) the greater is its amplitude. Pick up the best choice for the answer, and explain your reasoning. Answer: f=1/T, f is reciprocal of T, =cT=c/f Therefore, the lower the frequency of EM in vacuum, then the longer is its period, and the larger is its wave-length therefore d) and e) are both correct.

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Q6. The amplitude of the wave in the figure is: a) 6 m; b) 36 cm; c) 25 km; d) 4 miles; e) 81 nm; f) 4m. Pick up the best choice for the answer, and explain your reasoning. Answer: The amplitude is 4 m. The units are represented on y axis. Q7. The frequency of the wave is: a)16 Hz; b) 0.7 Hz; c) 5 m/s; d) 0.5 Hz; e) 2 Hz; f) 1 Hz; g) 4 Hz. Pick the best choice for the answer, and explain your reasoning. Answer: The frequency is 0.5 Hz. The are two periods in 4 seconds, on the y axis.

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Q8. What is the angle of the first minima (dark fringe) in the diffraction pattern produced by 600 nm light incident upon a 600 nm slit: a) 160 ; b) 120 ; c) 45 ; d) 90 ; e) 180 ; f) 0 ; g) 30 . Pick up the best choice for the answer, and explain your reasoning. Answer: The first minimum of diffraction through a single slit is given by the following relationship: a sin =.Then sin = /a=600/600 = 1 This value corresponds to =90 °. The correct answer is d).

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Q10. If the wavelength of an electromagnetic wave is 200 nm (2 10 -7 m) the wave is: a) a radio wave; b) a microwave; c) an infrared wave; d) a UV wave; e) a X-ray. Answer: The wavelength is lower than the blue limit (400 nm) of the visible region of the EM spectrum. Therefore, this is a UV wave (d). Q9. The figure represents an electromagnetic wave frozen in time. The upper horizontal line is: a) amplitude; b) frequency; c) period; d) wavelength; e) wave-speed. Pick up the best choice for the answer, and explain your reasoning. Answer: The distance between two maxima is a wavelength. Therefore, the answer is d).

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Q11. Let’s consider a double-slit interference experiment with the distance between the slits d=20 m (20 10 -6 m). The interference pattern is observed on a screen located at a distance L = 4 m from the slit. Calculate the distance between the first minima that correspond to red light ( =700 nm) and blue light ( =400 nm), respectively. Calculate the distance between the first maximum that corresponds to green light ( =550 nm) and the first maximum that corresponds to blue light ( =400 nm), respectively. Approximate sin tg . Answer: for m=0 (the first dark fringe) y dark = L/2d y dark (red)-y dark (blue)=(L/2d)×( red - blue )=[(4 m)/(2×20×10 -6 m)] ×300 nm= 10 5 ×3×10 -7 =3×10 -2 m=0.03 m For m=1 (the first bright fringe) y bright = L/2d; y bright (green)-y bright (blue)=(L/d)×( green - blue )=[(4 m)/(2×10 -5 m)]×150 nm= 2×10 5 ×1.5×10 -7 m=3×10 -2 m=0.03 m

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