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qz12sol_7193f10 - 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: λ is 40 cm. The speed of the wave is v = √ (F/ μ ) = 28.3 m/s. The angular frequency is ω = vk = v(2 π / λ ) = 445 rad/s The transverse velocity is the time derivative of the displacement. We get vt(x,t) = dy/dt = ω y m cos(kx - ω t + φ ) Cosine function changes from -1 to 1. Therefore, the maximum transverse velocity is v t,max = ω y m = 44.5 m/s b) Find the phase constant φ . The graph shows the displacement at t = 0. The wave equation at t = 0 is y(x,0) = y m sin(kx + φ ). At x = 0, the displacement is –y s . Therefore, the phase constant is y(0,0) - = y m sin( φ ) = –y s ⇒ φ = sin-1 (y s /y m ) = 233º or 307º 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( φ ) < 0 90º < φ < 270º The phase constant is 233º....
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