Physics Solution Manual for 1100 and 2101

# 3 n 028 m m l 2 130 hz 56 10 3 kgm chapter 17

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Unformatted text preview: when the difference in the distances traveled by the two sound waves in reaching point P is an integer number of wavelengths. That is, when (L – x) – x = nλ where n is an integer (or zero). Solving for x gives x= L − nλ 2 (1) When n = 0, x = L/2 = (7.80 m)/2 = 3.90 m . This corresponds to the point halfway between the two speakers. Clearly in this case, each wave has traveled the same distance and therefore, they will arrive in phase. When n = 1, x= (7.80 m) − (4.70 m) = 1.55 m 2 Thus, there is a point of constructive interference 1.55 m from speaker A . The points of constructive interference will occur symmetrically about the center point at L/2, so there is also a point of constructive interference 1.55 m from speaker B, that is at the point 7.80 m – 1.55 m = 6.25 m from speaker A . When n > 1, the values of x obtained from Equation (1) will be negative. These values correspond to positions of constructive interference that lie to the left of A or to the right of C. They do not lie on the line between the speakers. Chapter 17 Problems 911 12. REASONING Equation 17.1 specifies the diffraction angle θ according to sin θ = λ / D , where λ is the wavelength of the sound and D is the width of the opening. The wavelength depends on the speed and frequency of the sound. Since the frequency is the same in winter and summer, only the speed changes with the temperature. We can account for the effect of the temperature on the speed by assuming that the air behaves as an ideal gas, for which the speed of sound is proportional to the square root of the Kelvin temperature. SOLUTION Equation 17.1 indicates that sin θ = λ D Into this equation, we substitute λ = v / f (Equation 16.1), where v is the speed of sound and f is the frequency: λ v/ f sin θ = = D D Assuming that air behaves as an ideal gas, we can use v = γ kT / m (Equation 16.5), where γ is the ratio of the specific heat capacities at constant pressure and constant volume, k is Boltzmann’s constant, T is the Kelvin temperature, and m is the average mass of the molecules and atoms of which the air is composed: sin θ = v 1 γ kT = fD fD m Applying this result for each temperature gives sin θsummer = 1 γ kTsummer fD m and sin θ winter = 1 γ kTwinter fD m Dividing the summer-equation by the winter-equation, we find sin θsummer sin θ winter 1 γ kTsummer T fD m = = summer Twinter 1 γ kTwinter fD m Thus, it follows that sin θsummer = sin θ winter Tsummer Twinter = sin 15.0° 311 K = 0.276 273 K or θsummer = sin −1 ( 0.276 ) = 16.0° 912 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA 13. SSM REASONING The diffraction angle for the first minimum for a circular opening is given by Equation 17.2: sinθ = 1.22 λ / D , where D is the diameter of the opening. SOLUTION a. Using Equation 16.1, we must first find the wavelength of the 2.0-kHz tone: λ= v 343 m/s = = 0.17 m f 2.0 × 10 3 Hz The diffraction angle for a 2.0-kHz tone is, therefore, θ = sin –1 1.22 × 0.17 m = 0.30 m 44° b. The wavelength of a 6.0-kHz tone is λ= v 343 m/s = = 0.057 m f 6.0 × 103 Hz Therefore, if we wish to generate a 6.0-kHz tone whose diffraction angle is as wide as that for the 2.0-kHz tone in part (a), we will need a speaker of diameter D, where D= 1.22 λ (1.22)(0.057 m) = = 0.10 m sin θ sin 44° 14. REASONING The diffraction angle θ is determined by the ratio of the wavelength λ of the sound to the diameter D of the speaker, according to sin θ = 1.22 λ/D (Equation 17.2). The wavelength is related to the frequency f and the speed v of the wave by λ = v/f (Equation 16.1). SOLUTION Substituting Equation 16.1into Equation 17.2, we have sin θ = 1.22 λ D = 1.22 v Df Since the speed of sound is a constant, this result indicates that the diffraction angle θ will be the same for each of the three speakers, provided that the diameter D times the frequency f has the same value. Thus, we pair the diameter and the frequency as follows: Chapter 17 Problems 913 Diameter × Frequency = D f 3 2 (0.050 m)(12.0 × 10 Hz) = 6.0 × 10 m/s 3 2 (0.10 m)(6.0 × 10 Hz) = 6.0 × 10 m/s 3 2 (0.15 m)(4.0 × 10 Hz) = 6.0 × 10 m/s The common value of the diffraction angle, then, is θ = sin −1 1.22 v −1 1.22 ( 343 m/s ) = 44° = sin 2 Df 6.0 × 10 m / s 15. REASONING For a rectangular opening (“single slit”) such as a doorway, the diffraction angle θ at which the first minimum in the sound intensity occurs is given by sin θ = λ D (Equation 17.1), where λ is the wavelength of the sound and D is the width of the opening. This relation can be used to find the angle provided we realize that the wavelength λ is related to the speed v of sound and the frequency f by λ = v/f (Equation 16.1). SOLUTION a. Substituting λ = v/f into Equation 17.1 and using D = 0.700 m (only one door is open) gives λ v 343 m/s sin θ = = = = 0.807 θ = sin −1 ( 0.807 ) = 53.8° D f D ( 607 Hz )( 0.700 m ) b. When both doors are open, D = 2 × 0.700 m and the diffraction angle is sin...
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