527_Physics ProblemsTechnical Physics

527_Physics ProblemsTechnical Physics - P18.11 (a) Chapter...

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Chapter 18 529 P18.11 (a) φ 1 20 0 5 00 32 0 2 00 36 0 =− = ... . . rad cm cm rad s s rad bg af 1 25 0 5 00 40 0 2 00 45 0 900 516 156 = == ° = ° . . . rad cm cm rad s s rad radians (b) −−= + 20 0 32 0 25 0 40 0 5 00 8 00 .... .. xt At t = 200 . s , the requirement is φπ + = + 500 800200 2 1 . xn a f for any integer n . For x < 320 ., −+ 160 x is positive, so we have = + + 2 1 21 . . x n a f π , or The smallest positive value of x occurs for n = 2 and is x + = 41 00584 . . a f cm . P18.12 (a) First we calculate the wavelength: λ = v f 344 21 5 16 0 ms Hz m . . Then we note that the path difference equals 100 1 2 m m −= Therefore, the receiver will record a minimum in sound intensity. (b) We choose the origin at the midpoint between the speakers. If the receiver is located at point ( x , y ), then we must solve: xy ++ = 1 2 2 2 2 2 a f a f Then,
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