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qz12sol_7193f10

# qz12sol_7193f10 - Î is 40 cm The speed of the wave is v =...

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TA: Tomoyuki Nakayama Monday, December 6, 2010 PHY 2048: Physic 1, Discussion Section7193 Quiz 12 (Homework Set #14) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ A sinusoidal transverse wave is traveling along a string in the negative direction of an x axis. The wave equation is of the form y(x t) = y m sin( kx - ω t + φ ). The figure below right shows a plot of the displacement as a function of position at time t = 0; the y intercept is -8.0 cm. The string tension is 40 N, and its linear density is 50 g/m. a) What is the maximum transverse speed of a particle in the string? According to the graph, the maximum displacement y m is 10 cm and the wavelength
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Unformatted text preview: Î» is 40 cm. The speed of the wave is v = âˆš (F/ Î¼ ) = 28.3 m/s. The angular frequency is Ï‰ = vk = v(2 Ï€ / Î» ) = 445 rad/s The transverse velocity is the time derivative of the displacement. We get vt(x,t) = dy/dt = Ï‰ y m cos(kx - Ï‰ t + Ï† ) Cosine function changes from -1 to 1. Therefore, the maximum transverse velocity is v t,max = Ï‰ y m = 44.5 m/s b) Find the phase constant Ï† . The graph shows the displacement at t = 0. The wave equation at t = 0 is y(x,0) = y m sin(kx + Ï† ). At x = 0, the displacement is â€“y s . Therefore, the phase constant is y(0,0) - = y m sin( Ï† ) = â€“y s â‡’ Ï† = sin-1 (y s /y m ) = 233Âº or 307Âº Now we need to choose one of them. At x = 0, the slope of the graph is negative. This leads to dy/dx (0,0) = k y m cos( Ï† ) < 0 90Âº < Ï† < 270Âº The phase constant is 233Âº....
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