ch16 - CHAPTER 16 WAVE MOTION ActivPhysics can help with...

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Unformatted text preview: CHAPTER 16 WAVE MOTION ActivPhysics can help with these problems: Activities 10.1, 10.2, 10.7, 10.10 Section 16-2: Wave Properties Problem 1. Ocean waves with 18-m wavelength travel at 5 . 3 m / s . What is the time interval between wave crests passing under a boat moored at a fixed location? Solution Wave crests (adjacent wavefronts) take a time of one period to pass a fixed point, traveling at the wave speed (or phase velocity) for a distance of one wavelength. Thus T = /v = 18 m / (5 . 3 m / s) = 3 . 40 s . Problem 2. Ripples in a shallow puddle are propagating at 34 cm / s . If the wave frequency is 5 . 2 Hz , what are (a) the period and (b) the wavelength? Solution Equation 16-1 gives T = 1 /f = 1 / 5 . 2 Hz = 0 . 192 s and = v/f = vT = (34 cm / s) / (5 . 2 Hz) = 6 . 54 cm . Problem 3. An 88.7-MHz FM radio wave propagates at the speed of light. What is its wavelength? Solution From Equation 16-1, = v/f = (3 10 8 m / s) (88 . 7 10 6 Hz) = 3 . 38 m . Problem 4. One end of a rope is tied to a wall. You shake the other end with a frequency of 2.2 Hz, producing waves whose wavelength is 1.6 m. What is their propagation speed? Solution v = f = (2 . 2 Hz)(1 . 6 m) = 3 . 52 m / s (see Equa- tion 16-1). Problem 5. A 145-MHz radio signal propagates along a cable. Measurement shows that the wave crests are spaced 1.25 m apart. What is the speed of the waves on the cable? Compare with the speed of light in vacuum. Solution The distance between adjacent wave crests is one wavelength, so the wave speed in the cable (Equa- tion 16-1) is v = f = (145 10 6 Hz)(1 . 25 m) = 1 . 81 10 8 m / s = 0 . 604 c, where c = 3 10 8 m / s is the wave speed in vacuum. Problem 6. Calculate the wavelengths of (a) a 1.0-MHz AM radio wave, (b) a channel 9 TV signal (190 MHz), (c) a police radar (10 GHz), (d) infrared radiation from a hot stove (4 . 10 13 Hz) , (e) green light (6 . 10 14 Hz) , and (f) 1 . 10 18 Hz X rays. All are electromagnetic waves that propagate at 3 . 10 8 m / s . Solution With Equation 16-1 in an equivalent form, = vT = v/f, we find: (a) = (3 10 8 m / s) / (10 6 Hz) = 300 m , (b) = 1 . 58 m , (c) = 3 cm , (d) = 7 . 5 m, (e) = 500 nm, (f) = 3 A . (See Appendix C on units.) Problem 7. Detecting objects by reflecting waves off them is effective only for objects larger than about one wavelength. (a) What is the smallest object that can be seen with visible light (maximum frequency 7 . 5 10 14 Hz)? (b) What is the smallest object that can be detected with a medical ultrasound unit operating at 5 MHz? The speed of ultrasound waves in body tissue is about 1500 m/s. Solution (a) The wavelength of light corresponding to this maximum frequency is = c/f = (3 10 8 m / s) (7 . 5 10 14 Hz) = 400 nm , violet in hue (see Equa- tion 16-1). (b) The ultrasonic waves described have wavelength = v/f = (1500 m / s) / (5 MHz) = 0 . 3 mm ....
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This note was uploaded on 08/28/2009 for the course PHYS 2330 taught by Professor Wilke during the Spring '08 term at Rensselaer Polytechnic Institute.

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ch16 - CHAPTER 16 WAVE MOTION ActivPhysics can help with...

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