Solution_7_08_ - 1 Physics 214 Spring 2008 Solutions to...

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1 Physics 214 Spring 2008 Solutions to Assignment ] 7 Problem 1. ( a ). Given that ω 1 - ω 2 = 0 . 1 2 ( ω 1 + ω 2 ), we have ω 2 = 19 21 ω 1 . Therefore, s (0 , t ) = s m cos( ω 1 t ) + s m cos( ω 2 t ) = s m cos( ω 1 t ) + s m cos( 19 21 ω 1 t ). 0.02 0.04 0.06 0.08 t LParen1 s RParen1 Minus 2 Minus 1 1 2 s LParen1 cm RParen1 Figure 1: Plot of s (0 , t ) versus t for ω 1 = 1000 π rad/s and s m = 1 cm . ( b ). With k 2 = ω 2 v = 19 21 ω 1 v = 19 21 k 1 , we have s ( x, 0) = s m cos( k 1 t ) + s m cos( k 2 t ) = s m cos( k 1 t ) + s m cos( 19 21 k 1 t ). 5 10 15 20 25 x LParen1 m RParen1 Minus 2 Minus 1 1 2 s LParen1 cm RParen1 Figure 2: Plot of s ( x, 0) versus x for ω 1 = 1000 π rad/s which gives k 1 = ω 1 v 1000 π 343 m - 1 for sound wave in air with v 343 m/s and s m = 1 cm . ( c ). Yes. Because the wavelengths of the two sound waves are slightly different, there will be points where they are out of phase and destructively interfere as we see in figure 2. Problem 2. Given y ( x, t ) = A cos( kx - ωt ), we have dy ( x,t ) dx = - kA sin( kx - ωt ) and dy ( x,t ) dt = ωA sin( kx - ωt ) which give u p ( x, t ) = 1 2 τ ( dy ( x,t ) dx ) 2 = 1 2 τk 2 A 2 sin 2 ( kx - ωt ) and u k ( x, t ) = 1 2 μ ( dy ( x,t ) dx ) 2 = 1 2 μω 2 A 2 sin 2 ( kx - ωt ). Therefore, u p ( x, t ) = u k ( x, t ), since τk 2 = μω 2 or v = q τ μ = ω k . The total energy per unit length at time t is then u ( x, t ) = 2 u p ( x, t ) = 2 u k ( x, t ). Problem 3. No, the energies do not add up, since energy is not linear in y . The displacement waves add up. ( a ). The energy density of y = 2 f 1 ( x - vt ) is 4 times that of 2 f 1 ( x - vt ), since dy dx = 2 df 1 dx and dy dt = 2 df 1 dt and squaring these in calculating the potential and the kinetic energy densities (and consequently the total energy density) gives a factor of 4, and the total energy density in y is 4 times the energy density in f 1 .
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2 ( b ). For each non-overlapping pulse of triangular shape with width W and height A traveling to the right, we have df dx = 2 A W for the left half of the pulse and df dx = - 2
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