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Unformatted text preview: v = /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 + ). At x = 0, the displacement is zero. Thus the phase constant is y(0,0) = y m cos( ) = 0 = /2 or 3 /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( ) < 0 0 < < Therefore, the phase constant is /2. The wave number and angular frequency are, respectively, k = 2 / = 25.1 m-1, = 2 /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 + t + ) = 0.0148 m/s...
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This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.
- Summer '08