Physics Solution Manual for 1100 and 2101

# v is the speed of sound and l is the tube length we

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Unformatted text preview: then f2 < f1, and the f beat = f1 − f 2 = f1 − f 2 . Therefore, the frequency of the second beat frequency is carpenter’s hammering is f 2 = f1 − f beat , and we have that T2 = 1 1 1 1 = = = = 0.90 s 1 1 f 2 f1 − f beat 1 − 1 − T1 Tbeat 0.75 s 4.6 s 27. REASONING AND SOLUTION Since the wavelength is twice the distance between two successive nodes, we can use Equation 16.1 and see that v = λ f = (2L)f = 2(0.30 m)(4.0 Hz) = 2.4 m/s 28. REASONING According to Equation 17.3, the length L of the string is related to its third harmonic (n = 3) frequency f3 and the speed v of the waves on the string by v f3 = 3 2L or L= 3v 2 f3 (1) The speed v of the waves is found from Equation 16.2: v= F mL (16.2) Here, F is the tension in the string and the ratio m/L is its linear density. SOLUTION Substituting Equation 16.2 into Equation (1), then, gives the length of the string as: L= 3v 3 = 2 f3 2 f3 F mL (2) Although L appears on both sides of Equation (2), no further algebra is required. This is because L appears in the ratio m/L on the right side. This ratio is the linear density of the string, which has a known value of 5.6×10−3 kg/m. Therefore, the length of the string is L= 3 2 f3 F 3 3.3 N = = 0.28 m m L 2 (130 Hz ) 5.6 × 10 −3 kg/m Chapter 17 Problems 921 29. SSM REASONING The fundamental frequency f1 is given by Equation 17.3 with n = 1 : f1 = v /(2 L) . Since values for f1 and L are given in the problem statement, we can use this expression to find the speed of the waves on the cello string. Once the speed is known, the tension F in the cello string can be found by using Equation 16.2, v = F /(m / L) . SOLUTION Combining Equations 17.3 and 16.2 yields 2 L f1 = F m/ L Solving for F, we find that the tension in the cello string is F = 4 L2 f12 (m / L) = 4(0.800 m)2 (65.4 Hz)2 (1.56 ×10 –2 kg/m) = 171 N 30. REASONING The harmonic frequencies are integer multiples of the fundamental frequency. Therefore, for wire A (on which there is a second-harmonic standing wave), the fundamental frequency is one half of 660 Hz, or 330 Hz. Similarly, for wire B (on which there is a third-harmonic standing wave), the fundamental frequency is one third of 660 Hz, or 220 Hz. The fundamental frequency f1 is related to the length L of the wire and the speed v at which individual waves travel back and forth on the wire by f1 = v/(2L) (Equation 17.3, with n = 1). This relation will allow us to determine the speed of the wave on each wire. SOLUTION Using Equation 17.3 with n = 1, we find f1 = v 2L or v = 2 L f1 Wire A v = 2 (1.2 m )( 330 Hz ) = 790 m/s Wire B v = 2 (1.2 m )( 220 Hz ) = 530 m/s 31. SSM REASONING For standing waves on a string that is clamped at both ends, Equations 17.3 and 16.2 indicate that the standing wave frequencies are v fn = n 2L where v= F m/L Combining these two expressions, we have, with n = 1 for the fundamental frequency, 922 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA f1 = 1 F 2L m / L This expression can be used to find the ratio of the two fundamental frequencies. SOLUTION The ratio of the two fundamental frequencies is fold f new 1 Fold = 2L m / L = 1 Fnew 2L m / L Fold Fnew Since Fnew = 4 Fold , we have f new = f old Fnew Fold = f old 4 Fold Fold = f old 4 = (55.0 Hz) (2) = 1.10 ×102 Hz 32. REASONING Equation 17.3 (with n = 1) gives the fundamental frequency as f1 = v/(2L), where L is the wire’s length and v is the wave speed on the wire. The speed is given by F Equation 16.2 as v = , where F is the tension and m is the mass of the wire. m/ L SOLUTION Using Equations 17.3 and 16.2, we obtain v f1 = = 2L F m/L = 1 F = 1 2L 2 mL 2 160 N ( 6.0 ×10−3 kg ) ( 0.41 m ) = 130 Hz 33. REASONING A standing wave is composed of two oppositely traveling waves. The speed F v of these waves is given by v = (Equation 16.2), where F is the tension in the string m/L and m/L is its linear density (mass per unit length). Both F and m/L are given in the statement of the problem. The wavelength λ of the waves can be obtained by visually inspecting the standing wave pattern. The frequency of the waves is related to the speed of the waves and their wavelength by f = v/λ (Equation 16.1). Chapter 17 Problems 923 SOLUTION a. The speed of the waves is v= F 280 N = = 180 m/s m/ L 8.5 ×10−3 kg/m 1.8 m b. Two loops of any standing wave comprise one wavelength. Since the string is 1.8 m long and consists of three loops (see the drawing), the wavelength is λ= 2 3 (1.8 m ) = 1.2 m λ c. The frequency of the waves is f= v λ = 180 m/s = 150 Hz 1.2 m 34. REASONING The series of natural frequencies for a wire fixed at both ends is given by f n = nv / ( 2 L ) (Equation 17.3), where the harmonic number n takes on the integer values 1, 2, 3, etc.. This equation can be solved for n. The natural frequency fn is the lowest frequency that the human ear can detect, and the length L of the wire is given. The speed v at which waves travel on the wire can be obtained from the given values for the t...
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