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A transverse wave y = 6*mm*sin(100*pi*t - pi*x) travels along a string toward a boundary at x = 0. Beyond the boundary the linear mass density increases to four times its original value while the tension remains 100 N throughout. Determine which of the following statements about the reflected and transmitted waves are correct. (A) The reflected wave is y = -2*mm*sin(100*pi*t + pi*x) (B) The transmitted wave is y = 4*mm*sin(100*pi*t - 2*pi*x) (C) The average power carried by the incident wave is 1.8 W (D) The average power carried by the reflected wave is 0.6 W

1. The reflected wave is y = -2*mm*sin(100*pi*t + pi*x)
2. The transmitted wave is y = 4*mm*sin(100*pi*t - 2*pi*x)
3. The average power carried by the incident wave is 1.8 W
4. The average power carried by the reflected wave is 0.6 W

Correct answer: The reflected wave is y = -2*mm*sin(100*pi*t + pi*x)

Solution

From omega/k = 100*pi/pi = 100 m/s = v1. With T = 100 N and v1 = 100 m/s: mu1 = T/v1² = 0.01 kg/m. Since mu2 = 4*mu1 and T unchanged, v2 = sqrt(T/mu2) = 50 m/s. The reflection coefficient r = (50-100)/(50+100) = -1/3 gives amplitude -2 mm; transmission coefficient t = 2*50/150 = 2/3 gives amplitude 4 mm. The new wavenumber k2 = omega/v2 = 2*pi. Power of incident wave = (1/2)*mu1*v1*omega²*A² = 0.18*pi² ~ 1.78 W ~ 1.8 W. Power of reflected wave = (1/9)*1.8 ~ 0.2 W, not 0.6 W. So A, B, and C are correct.

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