Physics Solution Manual for 1100 and 2101

13 b to the left of the positive charge the two

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Unformatted text preview: 17.5, and the fundamental frequency f1 of the shortened tube is found by choosing n = 1. We note that the height h of the mercury column is equal to the difference between the original and final lengths of the air in the tube: h = L0 − L. SOLUTION Substituting h = L0 − L into P2 = P + ρ gh (Equation 11.4), we obtain atm P2 = Patm + ρ g ( L0 − L ) (1) The third harmonic frequency of the tube at its initial length L0 and the fundamental v frequency f1 of the tube at its final length L are equal, so from f n = n (Equation 17.5), 4L we find that v v f3,0 = 3 =1 =f 4L 4L 1 0 31 = L0 L or or L= L0 3 Substituting Equation (2) into Equation (1), we obtain ( ) 2 P2 = Patm + ρ g ( L0 − L ) = Patm + ρ g L0 − 1 L0 = Patm + 3 ρ gL0 3 Therefore, the pressure at the bottom of the mercury column is P2 = 1.01× 105 Pa + 2 (13600 kg/m3 ) ( 9.80 m/s 2 ) ( 0.75 m ) = 1.68 ×105 Pa 3 (2) 934 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA 51. REASONING The well is open at the top and closed at the bottom, so it can be approximated as a column of air that is open at only one end. According to Equation 17.5, the natural frequencies for such an air column are v fn = n 4L where n = 1, 3, 5, K The depth L of the well can be calculated from the speed of sound, v = 343 m/s, and a knowledge of the natural frequencies fn. SOLUTION We know that two of the natural frequencies are 42 and 70.0 Hz. The ratio of these two frequencies is 70.0 Hz 5 = 42 Hz 3 Therefore, the value of n for each frequency is n = 3 for the 42-Hz sound, and n = 5 for the 70.0-Hz sound. Using n = 3, for example, the depth of the well is L= nv 3 ( 343 m/s ) = = 6.1 m 4 f3 4( 42 Hz ) 52. REASONING AND SOLUTION The original tube has a fundamental given by f = v/(4L), so that its length is L = v/(4f ). The cut tube that has one end closed has a length of Lc = v/(4fc), while the cut tube that has both ends open has a length L0 = v/(2f0). We know that L = Lc + L0. Substituting the expressions for the lengths and solving for f gives fo fc ( 425 Hz ) ( 675 Hz ) = 162 Hz f= = 2 f c + f o 2 ( 675 Hz ) + 425 Hz 53. REASONING The time it takes for a wave to travel the length L of the string is t = L/v, where v is the speed of the wave. The speed can be obtained since the fundamental frequency is known, and Equation 17.3 (with n = 1) gives the fundamental frequency as f1 = v/(2L). The length is not needed, since it can be eliminated algebraically between this expression and the expression for the time. SOLUTION Solving Equation 17.3 for the speed gives v = 2Lf1. With this result for the speed, the time for a wave to travel the length of the string is Chapter 17 Problems t= 935 1 1 L L = = = = 1.95 ×10−3 s v 2 L f1 2 f1 2 ( 256 Hz ) 54. REASONING AND SOLUTION We know that L = v/(2f ). For 20.0 Hz L = (343 m/s)/[2(20.0 Hz)] = 8.6 m For 20.0 kHz L = (343 m/s)/[2(20.0 × 10 Hz)] = 8.6 × 10−3 m 3 55. SSM REASONING When constructive interference occurs again at point C, the path length difference is two wavelengths, or ∆s = 2 λ = 3.20 m . Therefore, we can write the expression for the path length difference as 2 2 sAC – sBC = sAB + sBC – sBC = 3.20 m This expression can be solved for sAB . SOLUTION Solving for sAB , we find that sAB = (3.20 m + 2.40 m) 2 – (2.40 m) 2 = 5.06 m 56. REASONING Let LA be length of the first pipe, and LB be the final length of the second pipe. The length ∆L removed from the second pipe, then, is ∆L = LA − LB (1) Both pipes are open at one end only, so their lengths L are related to their fundamental v frequencies f1 and the speed v of sound in air by f1 = (Equation 17.5 with n = 1), or 4L L= v 4 f1 (2) The beat frequency fbeat that occurs when both pipes are vibrating at their fundamental frequencies is the difference between the higher frequency f1,B of the second pipe and the lower frequency f1,A of the first pipe: 936 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA f beat = f1,B − f1,A (3) SOLUTION Substituting Equation (2) into Equation (1) yields ∆L = LA − LB = v v v 1 1 − = − 4 f1,A 4 f1,B 4 f1,A f1,B (4) Solving Equation (3) for the unknown frequency of the shortened pipe, we find that f1,B = f1,A + f beat (5) Substituting Equation (5) into Equation (4), we obtain ∆L = 343 m/s 1 v 1 1 v 1 1 1 − − − = = 4 f1,A f1,B 4 f1,A f1,A + f beat 256 Hz 256 Hz + 12 Hz 4 = 0.015 m Therefore, the second pipe has been shortened by cutting off cm (0.015 m ) 100m = 1.5 cm 1 57. REASONING The natural frequencies of a tube open at only one end are given by v Equation 17.5 as f n = n , where n is any odd integer (n = 1, 3, 5, …), v is the speed of 4L sound, and L is the length of the tube. We can use this relation to find the value for n for the 450-Hz sound and to determine the length of the pipe. SOLUTION v a. The frequency fn of the 450-Hz sound is given by 450 Hz = n . Likewise, the 4L v frequency of the next higher harmonic is 750 Hz = ( n + 2 ) , because n is an odd 4L integer and this means that the value of n for the next higher harmo...
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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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