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# chap 14 - WAVE MOTION 14 v= EXERCISES Section 14.1 Waves...

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14.1 WAVE MOTION E XERCISES Section 14.1 Waves and Their Properties 16. 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 = = = λ / ( . / ) . . 18 5 3 3 40 m/ m s s 17. I NTERPRET This problem is about wave propagation. Given the speed and frequency of the ripples, we are asked to compute the period and the wavelength. D EVELOP Equation 14.1 relates the speed of the wave to its period, frequency, and wavelength: v T f = = λ λ This is the equation we shall use to solve the problem. E VALUATE Equation 14.1 gives (a) T f = = = 1 1 5 2 0 192 . . Hz s, and (b) λ = = = v f 34 5 2 6 54 cm s Hz cm. / . . A SSESS The unit of frequency is Hz, with 1 1 1 Hz s = . If the frequency is kept fixed, then increasing the wavelength will increase the speed of propagation. 18. From Equation 14.1, λ = = × × = v f / ( / )/( . ) . . 3 10 88 7 10 3 38 8 6 m s Hz m 19. I NTERPRET This problem is about wave propagation. Given the speed and frequency of various electromagnetic waves, we are asked to compute their wavelength. D EVELOP Equation 14.1 relates the speed of the wave to its period, frequency, and wavelength: v T f v f = = = λ λ λ This is the equation we shall use to solve the problem. E VALUATE Since the speed of propagation of electromagnetic waves in vacuum is simply equal to the speed of light, v c = = × 3 0 10 8 . m/s, Equation 14.1 gives (a) λ = = = × c f 3 10 10 8 6 300 m s Hz m / ; (b) λ = = = × × c f 3 10 190 10 8 6 1 58 m s Hz m / . ; (c) λ = = = = × c f 3 10 10 8 10 0 03 3 m s Hz m cm / . ; (d) λ µ = = = × = × × c f 3 10 4 10 6 8 13 7 5 10 7 5 m s Hz m m / . . ; (e) λ = = = × = × × c f 3 10 6 10 7 8 14 5 0 10 500 m s Hz m nm / . ; (f ) λ = = = × = × × c f 3 10 1 0 10 10 8 18 3 0 10 3 m s Hz o m A / . . (See Appendix C on units.) A SSESS If the speed of propagation is kept fixed, then a higher frequency means a shorter wavelength. 20. The wave speed can be calculated from the distance and the travel time, which, together with the frequency and Equation 14.1, gives a wavelength of λ = = = × × = v f d t f / ( / )/ ( . ) . . 1200 5 60 3 1 1 29 km/ km 14

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14.2 Chapter 14 Section 14.2 Wave Math 21. I NTERPRET This problem is about the ultrasound wave. Given its frequency, and wavelength, we want to find its angular frequency, wave number, and wave speed. D EVELOP The relationships between the speed of the wave, its wave number, frequency, and wavelength are given by Equations 13.6, 14.1, and 14.2: f v T f k = = = = ω π λ λ π λ 2 2 , E VALUATE (a) Equation 13.6 gives ω π π = = = × 2 2 4 8 3 02 10 7 1 f ( . ) . . MHz s (b) Equation 14.2 gives k = = = × 2 2 0 31 4 1 2 03 10 π λ π . . . mm m (c) Using Equation 14.1, the speed of the ultrasound wave is v f = = × × = × λ ( . )( . ) . 4 8 10 0 31 10 1 49 10 6 3 3 Hz m m/s A SSESS The speed of the wave can also be computed as v k = = × × = × ω 3 02 10 2 03 10 1 49 10 7 1 4 1 3 . . . s m m/s Thus, we see that the pairs f , λ and ω , k are equivalent ways to describe the same wave.
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chap 14 - WAVE MOTION 14 v= EXERCISES Section 14.1 Waves...

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