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sol_hw02_Ch15

# sol_hw02_Ch15 - (c P = 4r2 I = 4(30.0 m 5 Y F 15.32(a Using...

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(c) P = 4 πr 2 I = 4 π (30 . 0 m ) 2 ( 10 . 0 W/m 2 ) = 110 kW . 5. Y & F 15.32 (a) Using the identity sin a + sin b = 2 sin 1 2 ( a + b ) cos 1 2 ( a - b ) , y net = y - + y + = A [sin ( kx + ωt ) + sin ( kx - ωt )] = 2 A sin 1 2 (2 kx ) cos 1 2 (2 ωt ) = 2 A sin ( kx ) cos ( ωt ) . X (b) We want to determine the location x * where y net ( x * , t ) = 0 for all t : y net ( x * , t ) = 2 A sin ( kx * ) cos ( ωt ) = 0 sin ( kx * ) = 0 kx * = x * = k = 2 π/λ = 2 . X 6. Y & F 15.37 If two functions y 1 and y 2 satisfy the wave equation, 1 v 2 2 ∂t 2 - 2 ∂x 2 y 1 = 1 v 2 2 ∂t 2 - 2 ∂x 2 y 2 = 0 , so does some linear combination/superposition of them: 1 v 2 2 ∂t 2 - 2 ∂x 2 ( A 1 y 1 + A 2 y 2 ) = A 1 1 v 2 2 ∂t 2 - 2 ∂x 2 y 1 + A 2 1 v 2 2 ∂t 2 - 2 ∂x 2 y 2 = 0 . The functions given, y 1 = A cos ( k 1 x - ω 1 t ) and y 2 = A cos ( k 2 x - ω 2 t ), are solutions of the wave equation provided ω 1 /k 1 = ω 2 /k 2 = v , which holds as stated in the problem. 7. Y & F 15.48 The ends of stick are free, so they must be antinodes. First harmonic: L = 1 2 λ 1 λ 1 = 2 L = 2 (2 . 0 m ) = 4 . 0 m . Second harmonic: L = λ 1 λ 1 = L = 2 . 0 m . Third harmonic: L = 3 2 λ 1 λ 1 = 2 3 L = 2 3 (2 . 0 m ) = 1 . 33 m . 8. Y & F 15.55 (a) ω = 2 πf = 2 π λ v = 2 π λ s F μ = r k 0 Δ m = s k 0 μ Δ x k 0 = Δ x 2 π λ 2 F . (b) k 0 is proportional to F , just as in Hooke’s law. It’s also independent of μ and A , just as it is for a simple harmonic oscillator. Finally, k 0 is inversely proportional to λ 2 , expressing the dependence on the curvature of the string. 9. Y & F 15.56 Summing torques about the pivot gives W L 2 cos θ = FL sin θ F = W 2 tan θ v = r FL m = r WL 2 m tan θ = 22 . 8 m/s
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sol_hw02_Ch15 - (c P = 4r2 I = 4(30.0 m 5 Y F 15.32(a Using...

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