{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

14_InstSolManual_PC

# 14_InstSolManual_PC - WAVE MOTION 14 v= EXERCISES Section...

This preview shows pages 1–3. Sign up to view the full content.

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 / 18 m/(5.3 m/s) 3.40 s. T v λ = = = 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 f T λ λ = = This is the equation we shall use to solve the problem. E VALUATE Equation 14.1 gives (a) 1 1 5.2 Hz 0.192 s, f T = = = and (b) 34 cm/s 5.2 Hz 6.54 cm. v f λ = = = A SSESS The unit of frequency is Hz, with 1 1 Hz 1s . - = If the frequency is kept fixed, then increasing the wavelength will increase the speed of propagation. 18. From Equation 14.1, 8 6 / (3 10 m/s)/(88.7 10 Hz) 3.38 m. v f λ = = × × = 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 v f T 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, 8 3.0 10 m/s, v c = = × Equation 14.1 gives (a) 8 6 3 10 m/s 10 Hz 300 m c f λ × = = = ; (b) 8 6 3 10 m/s 190 10 Hz 1.58 m c f λ × × = = = ; (c) 8 10 3 10 m/s 10 Hz 0.03 m 3 cm c f λ × = = = = ; (d) 8 13 3 10 m/s 6 4 10 Hz 7.5 10 m 7.5 m c f λ μ × - × = = = × = ; (e) 8 14 3 10 m/s 7 6 10 Hz 5.0 10 m 500 nm c f λ × - × = = = × = ; (f ) 8 18 o 3 10 m/s 10 1.0 10 Hz 3.0 10 m 3A c f λ × - × = = = × = (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 / ( / )/ 1200 km/(5 60 3.1) 1.29 km. v f d t f λ = = = × × = Section 14.2 Wave Math 14.1 14

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
14.2 Chapter 14 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: 2 , 2 f v f k T ϖ λ π λ π λ = = = = E VALUATE (a) Equation 13.6 gives 7 1 2 2 (4.8 MHz) 3.02 10 s . f ϖ π π - = = = × (b) Equation 14.2 gives 4 1 2 2 0.31 mm 2.03 10 m . k π π λ - = = = × (c) Using Equation 14.1, the speed of the ultrasound wave is 6 3 3 (4.8 10 Hz)(0.31 10 m) 1.49 10 m/s v f λ - = = × × = × A SSESS The speed of the wave can also be computed as 7 1 3 4 1 3.02 10 s 1.49 10 m/s 2.03 10 m v k ϖ - - × = = = × × Thus, we see that the pairs , f λ and , k ϖ are equivalent ways to describe the same wave. 22. From Equation 14.2, (a) 1 2 /10.8 m 0.582 m , k π - = = and (b) 1 2 /(4.1 s) 1.53 s . ϖ π - = = 23. I NTERPRET We are given a function that describes a traveling sinusoidal wave, and asked to compute various physical quantities associated with the wave.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}