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chap. 15

# chap. 15 - MECHANICAL WAVES 15 2 = v/f f remains 120 2 so...

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M ECHANICAL W AVES 15 15.15. IDENTIFY and SET UP: Use Eq.(15.13) to calculate the wave speed. Then use Eq.(15.1) to calculate the wavelength. EXECUTE: (a) The tension F in the rope is the weight of the hanging mass: 2 (1.50 kg)(9.80 m/s ) 14.7 N Fm g == = / 14.7 N/(0.0550 kg/m) 16.3 m/s vF μ = (b) vf λ = so / (16.3 m/s)/120 Hz 0.136 m. = (c) EVALUATE: /, = where . g = Doubling m increases v by a factor of 2. /. = f remains 120 Hz and v increases by a factor of 2, so increases by a factor of 15.17. IDENTIFY: For transverse waves on a string, / = . = . SET UP: The wire has / (0.0165 kg)/(0.750 m) 0.0220 kg/m mL = . EXECUTE: (a) . The tension is . 2 (875 Hz)(3.33 10 m) 29.1 m/s × = 22 (0.0220 kg/m)(29.1 m/s) 18.6 N Fv = (b) 29.1 m/s v = EVALUATE: If is kept fixed, the wave speed and the frequency increase when the tension is increased. 15.20. IDENTIFY: Apply Eq.(15.25). SET UP: 2 f ω π = . / = . EXECUTE: (a) av 1 2 PF μω = A . 3 23 2 av 13 . 0 0 1 0 k g (25.0 N)(2 (120.0 Hz)) (1.6 10 m) 0.223 W 20 . 8 0 m P ⎛⎞ × ⎜⎟ ⎝⎠ = or 0.22 W to two figures. (b) is proportional to av P 2 A , so halving the amplitude quarters the average power, to 0.056 W. EVALUATE: The average power is also proportional to the square of the frequency. 15.24. IDENTIFY: The tension and mass per unit length of the rope determine the wave speed. Compare given in the problem to the general form given in Eq.(15.8). (,) yxt / vk = . The average power is given by Eq. (15.25). SET UP: Comparison with Eq.(15.8) gives 2.33 mm A = , 6.98 rad/m k = and 742 rad/s = . EXECUTE: (a) 2.30 mm A = (b) 742 rad s 118 Hz 2 2 f = . (c) 0.90 m 6.98 rad m k = (d) 742 rad s 106 m s 6.98 rad m v k = (e) The wave is traveling in the x direction because the phase of has the form (, ) yx t . kx ω t +

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(f) The linear mass density is 33 (3.38 10 kg) (1.35 m) 2.504 10 kg/m μ −− = × , so the tension is 23 2 (2.504 10 kg m)(106.3 m s) 28.3 N Fv μ == × = . (g) 22 3 2 3 2 11 av (2.50 10 kg m)(28.3 N)(742 rad s) (2.30 10 m) 0.39 W PF A μω × × = EVALUATE: In part (d) we could also calculate the wave speed as vf λ = and we would obtain the same result.
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chap. 15 - MECHANICAL WAVES 15 2 = v/f f remains 120 2 so...

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