ch16-p057 - of the wave is v = ω k = 1.00 m/s. Therefore,...

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The anti-node moves through 12 cm in simple harmonic motion, just as a mass on a vertical spring would move from its upper turning point to its lower turning point – which occurs during a half-period. Since the period T is related to the angular frequency by Eq. 15-5, we have T = 2 π ω = 2 π 4.00 π = 0.500 s . Thus, in a time of t = 1 2 T = 0.250 s, the wave moves a distance Δ x = vt where the speed
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Unformatted text preview: of the wave is v = ω k = 1.00 m/s. Therefore, Δ x = (1.00 m/s)(0.250 s) = 0.250 m. 57. Recalling the discussion in section 16-12, we observe that this problem presents us with a standing wave condition with amplitude 12 cm. The angular wave number and frequency are noted by comparing the given waves with the form y = y m sin( k x ± ω t )....
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This note was uploaded on 04/16/2011 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.

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