Physics Solution Manual for 1100 and 2101

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Unformatted text preview: TION Substituting the tension F = Mg and the speed v = L/t into Equation 16.2 for the speed of the pulse on the string gives v= F m/ L L Mg = t m/ L or Solving for the acceleration g due to gravity, we obtain L t g= 2 2 ( ) 0.95 m −4 (m / L) 1.2 ×10 kg/m 0.016 s = = 7.7 m/s 2 M 0.055 kg 16. REASONING Each pulse travels a distance that is given by vt, where v is the wave speed and t is the travel time up to the point when they pass each other. The sum of the distances traveled by each pulse must equal the 50.0-m length of the wire, since each pulse starts out from opposite ends of the wires. SOLUTION Using vA and vB to denote the speeds on either wire, we have vA t + vB t = 50.0 m Solving for the time t and using Equation 16.2 v = t= 50.0 m = vA + v B 50.0 m FA m/L + FB m/L = F , we find m/ L 50.0 m 6.00 × 10 N + 0.020 kg/m 2 3.00 × 10 2 N 0.020 kg/m = 0.17 s Chapter 16 Problems 841 17. REASONING The speed v of a transverse wave on a string is given by v = F / ( m /L ) (Equation 16.2), where F is the tension and m/L is the mass per unit length (or linear density) of the string. The strings are identical, so they have the same mass per unit length. However, the tensions are different. In part (a) of the text drawing, the string supports the entire weight of the 26-N block, so the tension in the string is 26 N. In part (b), the block is supported by the part of the string on the left side of the middle pulley and the part of the string on the right side. Each part supports one-half of the block’s weight, or 13 N. Thus, the tension in the string is 13 N. SOLUTION a. The speed of the transverse wave in part (a) of the text drawing is v= F 26 N = = 2.0 × 101 m/s m /L 0.065 kg/m b. The speed of the transverse wave in part (b) of the drawing is v= F 13 N = = 1.4 × 101 m/s m /L 0.065 kg/m 18. REASONING a. The wavelength is the horizontal distance between two successive crests. The horizontal distance between successive crests of wave B is two times greater than that of wave A. Therefore, B has the greater wavelength. b. The frequency f of a wave is related to its speed v and wavelength λ by Equation 16.1, f = v/ λ. Since the speed is the same for both waves, the wave with the smaller wavelength has the larger frequency. Therefore wave A, having the smaller wavelength, has the larger frequency. c. The maximum speed of a particle attached to a wave is given by Equation 10.8 as vmax = Aω, where A is the amplitude of the wave and ω is the angular frequency, ω = 2π f. Wave A has both a larger amplitude and frequency, so the maximum particle speed is greater for A. SOLUTION a. From the drawing, we determine the wavelength of each wave to be λA = 2.0 m and λB = 4.0 m 842 WAVES AND SOUND b. The frequency of each wave is given by Equation 16.1 as: fA = v λA = 12 m/s = 6.0 Hz 2.0 m and f B = v λB = 12 m/s = 3.0 Hz 4.0 m c. The maximum speed for a particle moving in simple harmonic motion is given by Equation 10.8 as vmax = Aω. The amplitude of each wave can be obtained from the drawing: AA = 0.50 m and AB = 0.25 m. Wave A vmax = AAω A = AA 2π f A = ( 0.50 m ) 2π ( 6.0 Hz ) = 19 m/s Wave B vmax = ABω B = AB 2π f B = ( 0.25 m ) 2π ( 3.0 Hz ) = 4.7 m/s ______________________________________________________________________________ 19. SSM REASONING Newton's second law can be used to analyze the motion of the blocks using the methods developed in Chapter 4. We can thus determine an expression that relates the magnitude P of the pulling force to the magnitude F of the tension in the wire. Equation 16.2 [ v = F /( m / L) ] can then be used to find the tension in the wire. SOLUTION The following drawings show a schematic of the situation described in the problem and the free-body diagrams for each block, where m1 = 42.0 kg and m 2 = 19.0 kg . The pulling force is P, and the tension in the wire gives rise to the forces F and −F, which act on m1 and m2, respectively. Tension in wire FN1 m2 m1 F FN2 F –F P −FF m1g P m2g Newton's second law for block 1 is, taking forces that point to the right as positive, F = m1a , or a = F / m1 . For block 2, we obtain P − F = m2 a . Using the expression for a obtained from the equation for block 1, we have m P–F =F 2 m 1 or m P = F 2 +F = F m 1 m2 m + 1 1 According to Equation 16.2, F = v 2 (m / L ) , where m/L is the mass per unit length of the wire. Combining this expression for F with the expression for P, we have Chapter 16 Problems 843 m 19.0 kg P = v 2 (m / L) 2 + 1 = (352 m/s) 2 (8.50 ×10−4 kg/m) + 1 = 153 N m 42.0 kg 1 ______________________________________________________________________________ F mL (Equation 16.2), where F is the tension in the strand and m/L is the ratio of the mass m of the strand to its length L. The tension F is directly proportional to the spider’s mass M, because the tension force exerted on the spider by the silk is equal in magnitude to the spider’s weight Mg, where g is the magnitude of the acceleration due to gravity: 20. REASONING The sp...
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