ch01-solutions.pdf - Chapter 1 Introduction to Waves...

This preview shows page 1 out of 5 pages.

Unformatted text preview: Chapter 1: Introduction to Waves Solutions PHYS2405 Question 1 A jet airplane takes 2 hours to travel 1800 miles in a straight line. What is the average velocity of the jet? a. 1800 miles per hour. b. 0.0011 miles per hour. c. 600 miles per hour. d. 900 miles per hour. e. none of the above. Solution The average velocity of the jet is = Question 2 Spring 2019 1800 miles = 2 hours ⇒ = 900 miles = 900 mi/h. hour Is the plane in Question (1) supersonic (faster than the speed of sound)? a. Yes. b. No. Solution Page 5 in the text lists the “propagation velocity of sound in air is approximately” 760 mph. Since 900 mph is greater than 760 mph, the plain is supersonic. Question 3 A grandfather clock ticks once when the pendulum swings from left to right and once again on the swing from right to left. There are two ticks each second. What is the frequency of the pendulum motion in cycles per second (or Hertz, Hz)? a. 0.5 Hz. b. 1 Hz. c. 2 Hz. d. 4 Hz. e. none of the above. Solution Frequency is the number of cycles per second. Since there are two ticks each second, the pendulum completes one entire cycle each second — one ‘tick’ for left to right and another ‘tick’ for right to left. This means the period of the pendulum, the time for a complete cycle, is 1 second. So the frequency is cycle 1 1 = = seconds = 1 ⇒ = 1 Hz. 1 second cycle Question 4 What is the frequency of the clock in Question (3) in cycles per minute (cpm)? a. 20 cpm. b. 30 cpm. c. 60 cpm. d. 120 cpm. e. none of the above. Solution Rather than expressing the frequency in cycles per second, we need to express it as cycles per minute. Since the pendulum completes one cycle each second and there are 60 seconds per each minute, it completes 60 cycles in one minute. This is exactly the frequency: 60 cycles per minute or 60 cpm. Question 5 A swimmer standing near the edge of a lake notices a cork bobbing in the water. While watching for one minute, she notices the cork bob (from up to down to back up) 240 times. What is the frequency in Hz of the water wave going by? a. 0.5 Hz. b. 1 Hz. c. 4 Hz. d. 120 Hz. e. 240 Hz. Solution Since the cork bobs from up to down and up 240 times, the cork completes 240 cycles. This occurred during the one minute the swimmer watched the cork. So the frequency of the water wave is (remember that there are 60 seconds in one minute): 240 cycle cycles = ⇒ =4 = 4 Hz. sec 60 s Question 6 Which of the following best describes the water wave in Question (5)? a. longitudinal. b. oscillatory. c. impulsive. d. none of the above. Solution Water waves are transverse. Since the cork bobbed 240 times, the water wave was not an impulse but repetitive, or in other words: oscillatory. Stephens PHYS2405 Question 7 Chapter 1 Solutions, page 2 of 5 Spring 2019 What is the wave velocity of the water wave in Question (5) in feet per second (ft/s)? a. 0.5 ft/s. b. 1 ft/s. c. 60 ft/s. d. 120 ft/s. e. there is inadequate information to compute the velocity. Solution To calculate the wave velocity, we need wavelength and frequency . We only know so we are not provided with enough information about the wave to calculate the velocity. Question 8 The cork in Question (5) travels 1 foot in moving from its lowest position to its highest position. What is the average velocity of the cork during this motion? a. 1 ft/s. b. 2 ft/s. c. 3 ft/s. d. 4 ft/s. e. none of the above. Solution The cork traveled one foot during the time it took, starting from a crest, for an adjacent trough to pass. This corresponds to half a wavelength or half the water wave’s period. Since we found the frequency earlier to be 4 Hz, we can find the period as 1 1 cycle 1 sec 1 = = = = s. 2 cycle 4 4 sec 1 The time required for the cork to travel its one-foot is half of this: 8 s. So the average velocity of the cork is 1 ft = Question 9 = 1 s 8 ⇒ = 8 ft/s. The velocity you found in Question (8) is a. the average medium velocity. b. the average wave velocity. c. neither of these two terms. Solution Since the velocity dealt with the motion of the cork, and therefore of the water, and not the motion of a crest or trough, it must be the medium velocity and not the wave velocity. Question 10 A student sleeping in class snores 90 times in 3 minutes. What is the frequency of snoring in snores per second (sn/s)? a. 0.5 sn/s. b. 1 sn/s. c. 2 sn/s. Solution The frequency is = 90 snores 90 snores = 3 minutes 180 seconds d. 60 sn/s. ⇒ e. none of the above. = 0.5 sn/s. DISPLACEMENT Question 11 0 1 2 3 4 5 DISTANCE (ft) 6 7 8 Figure 1 A wave on a rope, at one particular instant in time, looks as shown in Figure 1. What is the wavelength of the wave? a. 1 ft. Stephens b. 2 ft. c. 4 ft. d. 5 ft. e. 8 ft. PHYS2405 Chapter 1 Solutions, page 3 of 5 Spring 2019 Solution The separation from the first crest to the second corresponds to a distance of 4 ft. Question 12 responsible) a. b. c. d. e. An explosion might be seen before it is heard because (pick the one option that provides the best answer that is undoubtedly sound waves have a smaller propagation velocity than light waves. there is constructive interference. sound gets fainter the farther you are from where it was created. friction slows down the sound wave. a, c, and d are all involved. Solution Sound travels at about 1100 ft/s whereas light travels at 186,000 ft/s. Whether there is interference or not is irrelevant. The sound does get fainter but that does not affect the wave’s velocity. Friction does not slow the wave; it attenuates it. So the only valid answer is (a). Question 13 Assume that the velocity of sound in air is about 1100 feet per second. A tuning fork that vibrates at 250 Hz gives rise to a sound wave having a wavelength of about a. 0.227 ft. b. 4.4 ft. c. 250 ft. d. 275,000 ft. Solution Using the formula = , we can perform some algebra to find 1100 fts = → = → == 250 1s Question 14 ⇒ = 4.4 ft. A pure tone is always a. an impulsive wave. b. electronically generated. c. at very low frequency. d. a sinusoidal sound wave. e. of small amplitude. Solution To be a tone, the wave must be periodic and not impulsive. We frequently use a tuning fork to produce a pure tone so it does not need to be electronically generated. It can have any frequency you want (as long as it is audible) and can have any amplitude you want (volume). So we only have one response remaining. Question 15 Which of the following is false? a. The sound waves from two tuning forks of the same frequency will interfere constructively at some points. b. Two waves, and , have the same wave velocity. If has a higher frequency than , will have a shorter wavelength. c. In a transverse wave, the motion of the medium is in the same direction as the wave velocity. Solution Given two sources of sound, the waves will interfere somewhere if they overlap. If the two waves have the same velocity , then our formula = guarantees that (b) is true. So we only have (c) to choose. Consider the two graphs in Figure 2, denoting the same traveling wave on a string. Questions 16–22 refer to these graphs. Question 16 a. 1 ft. The wavelength is b. 2 ft. c. 3 ft. Solution From the top graph, the wave completes a cycle in two feet. Stephens d. 6 ft. At t = 0 s At x = 0.5 ft Question 17 a. 1 s. DISPLACEMENT (ft) Chapter 1 Solutions, page 4 of 5 DISPLACEMENT (ft) PHYS2405 Spring 2019 2 1 0 −1 −2 0 1 2 3 x (feet) 4 5 6 2 1 0 −1 −2 0 1 2 3 4 5 t (seconds) 6 7 8 Figure 2 The period is b. 2 s. c. 3 s. d. 4 s. Solution From the lower graph, the wave complete a cycle in four seconds. Question 18 a. 1/8 Hz. The frequency is b. 1/4 Hz. Solution c. 1/2 Hz. d. 1 Hz. c. 3 ft. d. 6 ft. 1 Since the period = 4 s, = 1/ = 4 Hz. Question 19 a. 1 ft. The amplitude is b. 2 ft. Solution From either graph, the height of a crest above the equilibrium line is two feet. Question 20 a. 1/4 ft/s. The wave velocity is b. 1/2 ft/s. Solution c. 1 ft/s. d. 2 ft/s. 1 Knowing the wavelength = 2 ft and the frequency = 4 Hz, the wave velocity is = = ( 14 Hz) (2 ft) ⇒ = 12 ft/s. Question 21 a. −1 ft. The height of the string at = 1 ft and = 0 s is b. −2 ft. c. 0 ft. d. 1 ft. e. 2 ft. Solution The height of the wave at the given position and time is determined from the top graph, since it shows the displacement of the wave at = 0 and for various values of . At one foot corresponds to zero displacement. Stephens PHYS2405 Question 22 a. −1 ft. Chapter 1 Solutions, page 5 of 5 Spring 2019 The height of the string at = 1 ft and = 4 s is b. −2 ft. c. 0 ft. d. 1 ft. e. 2 ft. Solution The question requires a bit more thought than the previous questions. Neither of our graphs correspond to the time/position in question. But we know that the wave has a period of four seconds. This means that the top graph would look exactly the same if we changed the description from “At = 0 s” to “At = 4 s.” Reading the displacement at = 1 ft from the top graph, we find zero. Stephens ...
View Full Document

  • Spring '19
  • STEPHENS
  • Wavelength, Stephens, Hz of the water wave

{[ snackBarMessage ]}

Get FREE access by uploading your study materials

Upload your study materials now and get free access to over 25 million documents.

Upload now for FREE access Or pay now for instant access
Christopher Reinemann
"Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

Ask a question for free

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern