Download presentation

Presentation is loading. Please wait.

Published byKenneth Molde Modified over 2 years ago

1
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

2
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

3
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.

4
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.

5
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).

6
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).

7
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

Similar presentations

Presentation is loading. Please wait....

OK

Vibrations & Waves Chapter 25 - This will be phun!

Vibrations & Waves Chapter 25 - This will be phun!

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on working of human eye and defects of vision and their correction Ppt on video teleconferencing installation Ppt on cse related topics in biology Ppt on human chromosomes disorders Ppt on nepali culture and tradition Ppt on red blood cells Ppt on soft skills download Ppt on standing order act packets Ppt on blood stain pattern analysis publications Ppt on event driven programming examples