Physics Solution Manual for 1100 and 2101

# Ssm reasoning according to equation 1610 the sound

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Unformatted text preview: towards the car at this instant. Therefore, to calculate the Doppler-shifted vSE frequency of the horn blast, we must consider only the component vSE of the train’s velocity vtrain that is directed southeast. We will use trigonometry to find the magnitude vSE of this component of the train’s velocity, Car and then employ Equation 16.11 to determine the frequency fo heard by the driver. SOLUTION According to the drawing, the scalar component vSE of the train’s velocity vtrain that is directed towards the car is vSE = vtrain cos 45.0o Therefore, from Equation 16.11, we have that 1 f o = fs v 1 − SE v 1 = f o s 1 − vtrain cos 45.0 v 1 = ( 289 Hz ) ( 55.0 m/s ) cos 45.0o 1 − 343 m/s = 326 Hz 83. REASONING The Doppler shift that occurs here arises because both the source and the observer of the sound are moving. Therefore, the expression for the Doppler-shifted observed frequency fo is given by Equation 16.15 as 1 ± vo / v f o = fs 1m v / v s 884 WAVES AND SOUND where fs is the frequency emitted by the source, vo is the speed of the observer, vs is the speed of the source, and v is the speed of sound. The observer is moving toward the source, so we use the plus sign in the numerator. The source is moving toward the observer, so we use the minus sign in the denominator. Thus, Equation 16.15 becomes 1 + vo / v f o = fs 1− v / v s Recognizing that both trucks move at the same speed, we can substitute vo = vs = vTruck and solve for vTruck. SOLUTION Using Equation 16.15 as described in the REASONING and substituting vo = vs = vTruck, we have 1 + vTruck / v fo = fs 1− v Truck / v f v − o Truck fs f s v fo or vTruck = 1+ v Rearranging, with a view toward solving for vTruck/v, gives fo fs −1 = vTruck v f v + o Truck f v s or vTruck f o fo −1 1 + = v fs fs Finally, we obtain vTruck v fo −1 fs 1.14 − 1 0.14 = = = f o 1 + 1.14 2.14 1+ fs or 0.14 vTruck = ( 343 m/s ) = 22 m/s 2.14 84. REASONING a. Since the sound source and the observer are stationary, there is no Doppler effect. The wavelength remains the same and the frequency of the sound heard by the observer remains the same as that emitted by the sound source. b. When the sound source moves toward a stationary observer, the wavelength decreases (see Figure 16.28b). This decrease arises because the condensations “bunch-up” as the source moves toward the observer. The frequency heard by the observer increases, because, according to Equation 16.1, the frequency is inversely proportional to the wavelength; a smaller wavelength gives rise to a greater frequency. Chapter 16 Problems 885 c. The wavelength decreases for the same reason given in part (b). The increase in frequency is due to two effects; the decrease in wavelength, and the fact that the observer intercepts more wave cycles per second as she moves toward the sound source. SOLUTION a. The frequency of the sound is the same as that emitted by the siren; f o = fs = 2450 Hz The wavelength is given by Equation 16.1 as λ= 343 m /s v = = 0.140 m fs 2450 Hz b. According to the discussion in Section 16.9 (see the subsection “Moving source”) the wavelength λ ′ of the sound is given by λ ′ = λ − vs T , where vs is the speed of the source and T is the period of the sound. However, T = 1/fs so that λ ′ = λ − vs T = λ − vs fs = 0.140 m − 26.8 m /s = 0.129 m 2450 Hz The frequency fo heard by the observer is equal to the speed of sound v divided by the shortened wavelength λ ′ : v 343 m /s fo = = = 2660 Hz λ ′ 0.129 m c. The wavelength is the same as that in part (b), so λ ′ = 0.129 m . The frequency heard by the observer can be obtained from Equation 16.15, where we use the fact that the observer is moving toward the sound source: 14.0 m/s v 1+ o 1 + 343 m/s v = ( 2450 Hz ) = 2770 Hz f o = fs vs 1 − 26.8 m/s 1− 343 m/s v ______________________________________________________________________________ 85. REASONING Since the car is accelerating, its velocity is changing. The acceleration as of the car (the “source”) is given by Equation 2.4 as as = vs, final − vs, initial t (1) 886 WAVES AND SOUND where vs, final and vs, initial are the final and initial velocities of the source and t is the elapsed time. To find vs, final and vs, initial we note that the frequency fo of the sound heard by the observer depends on the speed vs of the source. When the source is moving away from a stationary observer, Equation 16.12 gives this relation as 1 f o = fs v 1+ s v where fs is the frequency of the sound emitted by the source and v is the speed of sound. Solving for vs gives f vs = v s − 1 fo (2) We assume that the car is accelerating along the +x axis, so its velocity is always positive. Since the speed of the car is the magnitude of the velocity, the speed has the same numerical value as the velocity. According to Equation (2), the final and initial velocities of the car are: [Final velocity] [Initial velocity] fs vs, final = v − 1 fo, final (3) fs vs, initial = v −1 f o,...
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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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