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Unformatted text preview: v = Î» /T = 0/0125 m/s b) What is the transverse velocity of the particle at x = 10.0 cm when t = 5.00 s? The graph shows the displacement of a string particle at t = 0. At t = 0, the wave equation is y(x,0) = y m cos(kx + Ï† ). At x = 0, the displacement is zero. Thus the phase constant is y(0,0) = y m cos( Ï† ) = 0 â‡’ Ï† = Ï€ /2 or 3 Ï€ /2 Now we need to choose one of the solutions. We use the fact that the slope of the graph is negative at x = 0. This leads to dy/dt(0.0) = k y m sin( Ï† ) < 0 0 < Ï† < Ï€ Therefore, the phase constant is Ï€ /2. The wave number and angular frequency are, respectively, k = 2 Ï€ / Î» = 25.1 m1, Ï‰ = 2 Ï€ /T = 0.314 s The transverse velocity at x = 10 cm, t = 5 s is v t (x = 0.1 m, t = 5 s) = dy/dt = ky m sin(kx + Ï‰ t + Ï† ) = 0.0148 m/s...
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 Summer '08
 Field
 Work, 10 cm, 0.1 m, 8.00 cm, Tomoyuki Nakayama, ym cos

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