problem16_76

University Physics with Modern Physics with Mastering Physics (11th Edition)

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16.76: a) b) From Eq. (16.4), the function that has the given 0 at ) 0 , ( = t x p is given graphically as shown. Each section is a parabola, not a portion of a sine curve. The period is 4 10 81 . 5 ) s m (344 m) 200 . 0 ( - × = = λ v s and the amplitude is equal to the area under the x p - curve between 0500 . 0 and 0 = = x x m divided by B , or m. 10 04 . 7 6 - × c) Assuming a wave moving in the x + -direction, ) , 0 ( t y is as shown. d) The maximum velocity of a particle occurs when a particle is moving throughout the origin, and the particle speed is . B pv x y y v v = - = The maximum velocity is found from the maximum pressure, and . s cm 9.69 Pa) 10 42 . 1 ( s) m Pa)(344
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Unformatted text preview: 40 ( 5 max = = y v The maximum acceleration is the maximum pressure gradient divided by the density, . s m 10 67 . 6 ) m kg 20 . 1 ( m) (0.100 Pa) . 80 ( 2 2 3 max = = a e) The speaker cone moves with the displacement as found in part (c ); the speaker cone alternates between moving forward and backward with constant magnitude of acceleration (but changing sign). The acceleration as a function of time is a square wave with amplitude 2 s m 667 and frequency kHz. 1.72 m) (0.200 ) s m 344 ( = = = v f...
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problem16_76 - 40 ( 5 max = = y v The maximum acceleration...

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