hw02_sol - 5 Let y1 = 2.0 mm(corresponding to time t1 and...

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5. Let y 1 = 2.0 mm (corresponding to time t 1 ) and y 2 = –2.0 mm (corresponding to time t 2 ). Then we find kx + 600 t 1 + φ = sin 1 (2.0/6.0) and kx + 600 t 2 + = sin 1 (–2.0/6.0) . Subtracting equations gives 600( t 1 t 2 ) = sin 1 (2.0/6.0) – sin 1 (–2.0/6.0). Thus we find t 1 t 2 = 0.011 s (or 1.1 ms).
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10. With length in centimeters and time in seconds, we have u = du dt = 225 π sin ( π x 15 π t ) . Squaring this and adding it to the square of 15 π y , we have u 2 + (15 π y ) 2 = (225 ) 2 [sin 2 ( x 15 t ) + cos 2 ( x 15 t )] so that u = (225 π ) 2 - (15 π y ) 2 = 15 π 15 2 - y 2 . Therefore, where y = 12, u must be ± 135 π . Consequently, the speed there is 424 cm/s = 4.24 m/s.
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(d) We choose the minus sign (between kx and ω t ) in the argument of the sine function because the wave is shown traveling to the right [in the + x direction] – see section 16-5). Therefore, with SI units understood, we obtain y = y m sin( kx kvt ) 0.0030 sin(16 x 2.4 ¡ 10 2 t ) . 11. (a) The amplitude y m is half of the 6.00 mm vertical range shown in the figure, i.e., 3.0 mm. m y = (b) The speed of the wave is v = d/t = 15 m/s, where d = 0.060 m and t = 0.0040 s. The angular wave number is k = 2 π/λ where λ = 0.40 m. Thus, k = 2 π λ = 16 rad/m . (c) The angular frequency is found from ω = kv = (16 rad/m)(15 m/s)=2.4 ¡ 10 2 rad/s.
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22. (a) The general expression for y ( x, t ) for the wave is y ( x, t ) = y m sin( kx ω t ), which, at x
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This note was uploaded on 04/05/2011 for the course PHYSICS 133 taught by Professor Steigman during the Fall '09 term at Ohio State.

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hw02_sol - 5 Let y1 = 2.0 mm(corresponding to time t1 and...

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