qz12sol_3891f10 - v = /T = 0/0125 m/s b) What is the...

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TA: Tomoyuki Nakayama Tuesday, December 7, 2010 PHY 2048: Physic 1, Discussion Section 3891 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 of wavelength 25.0 cm travels along a string in the negative direction of an x axis. The displacement y of the string particle at x = 0 is given in the figure below as a function of time t . The scale of the vertical axis is set by y s = 8.00 cm. The wave equation is to be in the form of y = y m cos( kx + ω t + φ ). a) What is the wave speed (speed of the wave along the string)? According to the graph, the period of the oscillation is 20s. The wave speed is
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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.

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qz12sol_3891f10 - v = /T = 0/0125 m/s b) What is the...

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