problem15_72

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

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  • PresidentHackerCaribou10582
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15.72: a) m 800 . 0 ) Hz 0 . 240 ( s) m 0 . 192 ( = = = λ f v , and the wave amplitude is cm. 400 . 0 SW = Α The amplitude of the motion at the given points is (i) 2) sin( cm) (0.400 (ii) node), (a 0 ) ( cm)sin 400 . 0 ( π π = antinode) (an cm 0.004 = cm. 0.283 4) sin( cm) (0.400 (iii) and = π b). The time is half of the period, or s. 10 08 . 2 ) 2 ( 1 3 - × = f c) In each case, the maximum velocity is the amplitude
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  • Wavelength, wave amplitude

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