F12P2CChap16Solutions

# D the two waves are 3 1 1 1 50 10 m sin 314m 314s y x

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(d) The two waves are ( ) ( ) ( ) 3 1 1 1 5.0 10 m sin 3.14m 314s y x t Ρਟ Τ੏ = × Σਿ Φ੯ and ( ) ( ) ( ) 3 1 1 2 5.0 10 m sin 3.14m 314s . y x t Ρਟ Τ੏ = × + Σਿ Φ੯ Thus, if one of the waves has the form ( , ) sin( ) m y x t y kx t ω = + , then the other wave must have the form ( , ) sin( ) m y x t y kx t ω ʹஒ = . The sign in front of ω for '( , ) y x t is minus.

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S12 Physics 2C CHAPTER 16 Solutions 53. (a) The amplitude of each of the traveling waves is half the maximum displacement of the string when the standing wave is present, or 0.25 cm. (b) Each traveling wave has an angular frequency of ω = 40 π rad/s and an angular wave number of k = π /3 cm –1 . The wave speed is v = ω / k = (40 π rad/s)/( π /3 cm –1 ) = 1.2 × 10 2 cm/s. (c) The distance between nodes is half a wavelength: d = λ /2 = π / k = π /( π /3 cm –1 ) = 3.0 cm. Here 2 π / k was substituted for λ . (d) The string speed is given by u ( x, t ) = y / t = – ω y m sin( kx )sin( ω t ). For the given coordinate and time, ( ) 1 1 9 (40 rad/s) (0.50cm) sin cm (1.5cm) sin 40 s s 0. 3 8 u Ρਟ Τ੏ Ρਟ Τ੏ π Λি Ξ৯ Λি Ξ৯ = π π = Μ৏ Ο৿ Μ৏ Ο৿ ΢ਯ Υ੟ ΢ਯ Υ੟ Νয় Πਏ Νয় Πਏ Σਿ Φ੯ Σਿ Φ੯ 55. Recalling the discussion in section 16-12, we observe that this problem presents us with a standing wave condition with amplitude 12 cm. The angular wave number and frequency are noted by comparing the given waves with the form y = y m sin( k x ± ω t ). The anti-node moves through 12 cm in simple harmonic motion, just as a mass on a vertical spring would move from its upper turning point to its lower turning point, which occurs during a half-period. Since the period T is related to the angular frequency by Eq. 15-5, we have 2 2 0.500 s. 4.00 T π π ω π = = = Thus, in a time of t = 1 2 T = 0.250 s, the wave moves a distance Δ x = vt where the speed of the wave is / 1.00 m/s. v k ω = = Therefore, Δ x = (1.00 m/s)(0.250 s) = 0.250 m.
S12 Physics 2C CHAPTER 16 Solutions 58. With the string fixed on both ends, using Eq. 16-66 and Eq. 16-26, the resonant frequencies can be written as , 1,2,3, 2 2 2 nv n n mg f n L L L τ μ μ = = = = K (a) The mass that allows the oscillator to set up the 4th harmonic ( 4 n = ) on the string is 2 2 2 2 2 2 2 4 4 4(1.20 m) (120 Hz) (0.00160 kg/m) 0.846 kg (4) (9.80 m/s ) n L f m n g μ = = = = (b) If the mass of the block is 1.00 kg m = , the corresponding n is 2 2 2 2 2 4 4(1.20 m) (120 Hz) (0.00160 kg/m) 3.68 9.80 m/s L f n g μ = = = which is not an integer. Therefore, the mass cannot set up a standing wave on the string.

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S12 Physics 2C CHAPTER 16 Solutions Hard 24. (a) The tension in each string is given by τ = Mg /2. Thus, the wave speed in string 1 is 2 1 1 1 (500g)(9.80m/s ) 28.6m/s. 2 2(3.00g/m) Mg v τ μ μ = = = = (b) And the wave speed in string 2 is 2 2 2 (500g)(9.80m/s ) 22.1m/s. 2 2(5.00g/m) Mg v μ = = = (c) Let 1 1 1 2 2 2 /(2 ) /(2 ) v M g v M g = = = μ μ and M 1 + M 2 = M . We solve for M 1 and obtain 1 2 1 500g 187.5g 188g. 1 / 1 5.00/3.00 M M μ μ = = = + + (d) And we solve for the second mass: M 2 = M M 1 = (500 g – 187.5 g) 313 g.
S12 Physics 2C CHAPTER 16 Solutions 59. (a) The frequency of the wave is the same for both sections of the wire. The wave speed and wavelength, however, are both different in different sections. Suppose there are n 1 loops in the aluminum section of the wire. Then, L 1 = n 1 λ 1 /2 = n 1 v 1 /2 f , where λ 1 is the wavelength and v 1 is the wave speed in that section. In this consideration,