sol_hw02_Ch15

sol_hw02_Ch15 - (c) P = 4 r 2 I = 4 (30 . m ) 2 ( 10 . W/m...

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Unformatted text preview: (c) P = 4 r 2 I = 4 (30 . m ) 2 ( 10 . W/m 2 ) = 110 kW . 5. Y & F 15.32 (a) Using the identity sin a + sin b = 2sin 1 2 ( a + b ) cos 1 2 ( a- b ) , y net = y- + y + = A [sin( kx + t ) + sin( kx- t )] = 2 A sin 1 2 (2 kx ) cos 1 2 (2 t ) = 2 A sin( kx )cos( t ) . 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( t ) = 0 sin( kx * ) = 0 kx * = n x * = n k = n 2 / = n 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- 1 t ) and y 2 = A cos( k 2 x- 2 t ), are solutions of the wave equation provided 1 /k 1 = 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 1 1 = 2 L = 2(2 . m ) = 4 . m . Second harmonic: L = 1 1 = L = 2 . m . Third harmonic: L = 3 2 1 1 = 2 3 L = 2 3 (2 . m ) = 1 . 33 m . 8. Y & F 15.55 (a) = 2 f = 2 v = 2 s F = r k m = s k x k = x 2 2 F . (b) k is proportional to F , just as in Hookes law. Its also independent of and A , just as it is for a simple harmonic oscillator. Finally, k is inversely proportional to 2 , expressing the dependence on the curvature of the string....
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sol_hw02_Ch15 - (c) P = 4 r 2 I = 4 (30 . m ) 2 ( 10 . W/m...

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