Physics Solution Manual for 1100 and 2101

# Substituting equation 2 for the volume v of the silk

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Unformatted text preview: s speed v relative to the car is equal to the car’s speed v relative to the road. The frequency f of the wave is equal to the frequency of the wheel vibrations caused by the wave. The distance between the centers of adjacent grooves, then, is the wavelength λ of the rumble-strip wave. The relationship among these three quantities is given by v = f λ (Equation 16.1), which we will use to find λ. SOLUTION Solving v = f λ (Equation 16.1) for the wavelength λ, we obtain λ= 3. v 23 m/s = = 0.28 m f 82 Hz REASONING a. The period is the time required for one complete cycle of the wave to pass. The period is also the time for two successive crests to pass the person. b. The frequency is the reciprocal of the period, according to Equation 10.5. c. The wavelength is the horizontal length of one cycle of the wave, or the horizontal distance between two successive crests. d. The speed of the wave is equal to its frequency times its wavelength (see Equation 16.1). e. The amplitude A of a wave is the maximum excursion of a water particle from the particle’s undisturbed position. Chapter 16 Problems 835 SOLUTION a. After the initial crest passes, 5 additional crests pass in a time of 50.0 s. The period T of the wave is 50.0 s T= = 10.0 s 5 b. Since the frequency f and period T are related by f = 1/T (Equation 10.5), we have f= 1 1 = = 0.100 Hz T 10.0 s c. The horizontal distance between two successive crests is given as 32 m. This is also the wavelength λ of the wave, so λ = 32 m d. According to Equation 16.1, the speed v of the wave is v = f λ = ( 0.100 Hz ) ( 32 m ) = 3.2 m/s e. There is no information given, either directly or indirectly, about the amplitude of the wave. Therefore, it is not possible to determine the amplitude. ____________________________________________________________________________________________ 4. REASONING The speed of a Tsunamis is equal to the distance x it travels divided by the time t it takes for the wave to travel that distance. The frequency f of the wave is equal to its speed divided by the wavelength λ, f = v/λ (Equation 16.1). The period T of the wave is related to its frequency by Equation 10.5, T = 1/f. SOLUTION a. The speed of the wave is (in m/s) v= x 3700 × 103 m 1 h = = 190 m/s t 5.3 h 3600 s b. The frequency of the wave is f= v λ = 190 m /s = 2.5 × 10−4 Hz 3 750 × 10 m (16.1) 836 WAVES AND SOUND c. The period of any wave is the reciprocal of its frequency: 1 1 = = 4.0 × 103 s (10.5) −4 f 2.5 × 10 Hz ______________________________________________________________________________ T= 5. SSM REASONING As the transverse wave propagates, the colored dot moves up and down in simple harmonic motion with a frequency of 5.0 Hz. The amplitude (1.3 cm) is the magnitude of the maximum displacement of the dot from its equilibrium position. SOLUTION The period T of the simple harmonic motion of the dot is T = 1 / f = 1/(5.0 Hz) = 0.20 s . In one period the dot travels through one complete cycle of its motion, and covers a vertical distance of 4 × (1.3 cm) = 5.2 cm . Therefore, in 3.0 s the dot will have traveled a total vertical distance of 3.0 s (5.2 cm) = 78 cm 0.20 s ______________________________________________________________________________ 6. REASONING AND SOLUTION The period of the wave is the same as the period of the person, so T = 5.00 s. a. f= 1 1 = = 0.200 Hz T 5.00 s (10.5) b. v = λ f = (20.0 m)(0.200 Hz) = 4.00 m/s (16.1) ______________________________________________________________________________ 7. REASONING The speed v of a wave is equal to its frequency f times its wavelength λ (Equation 16.1). The wavelength is the horizontal length of one cycle of the wave. From the left graph in the text it can be seen that this distance is 0.040 m. The frequency is the reciprocal of the period, according to Equation 10.5, and the period is the time required for one complete cycle of the wave to pass. From the right graph in the text, it can be seen that the period is 0.020 s, so the frequency is 1/(0.020 s). SOLUTION Since the wavelength is λ = 0.040 m and the period is T = 0.020 s, the speed of the wave is 1 1 v = λ f = λ = (0.040 m) = 0.20 m/s T 0.20 s _____________________________________________________________________________________________ Chapter 16 Problems 8. 837 REASONING The waves are traveling at a velocity vWS relative to the shore, and the jetskier is traveling in the same direction as the waves at a velocity vJS relative to the shore. Therefore, as explained in Section 3.4, the velocity of the jetskier relative to the shore is given by v JS = v JW + v WS (1) where vJW is the velocity of the jetskier relative to the waves. From the jetskier’s point of view, the waves have a speed that is the magnitude of vJW (or vJW), a wavelength λ equal to the distance between crests, and a frequency f equal to the frequency of the bumps he experiences. These three quantities are related by Equation 16.1: vJW = f λ (2) SOLUTION Solving Equation (1) for vJW...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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