A snapshot at time t = 0 of a transverse simusoidal traveling wave moving on a s

Question

A snapshot at time t = 0 of a transverse simusoidal traveling wave moving on a stretched string in the x direction with speed v = 125 m/s is shown in the figure. The tension in the string is F = 400 N. Assume that you can read the graph to three significant figures. What arc the amplitude A. wavelength lambda wave number k. angular frequency omega frequency f, and period T of the wave What is the linear mass density of the string? Write down the wave function for the transverse string wave versus both x and t. Obtain an expression for the transverse velocities of the particles in the string versus x and t and another expression for transverse accelerations of the particles in the string versus x and t. derived from the wave function in part (c). What is the maximum transverse speed and the maximum magnitude of the transverse acceleration of a particle in the string versus time In the figure, at what values of x is the linear kinetic energy density of the string maximum? Zero? At what values of x is the linear potential energy density of the string maximum? Zero? Explain. How much mechanical energy (kinetic plus potential energy) is present in one wavelength of the wave? What is the average power transported along the string by the wave? What is the original source of this power?

Explanation / Answer

! = 2f = 10 Note that the units of angular frequency are radians/second, not Hz. b) (5 pts) Compute the wave number k. v = f 20 = 5 = 4 m Note that we are not talking about an electro-magnetic wave here, so the speed is not 3×108 m/s! Then, using the definition of k, we have k = 2 = 2 4 = 2 c) (5 pts) Write an expression for the wave function of this wave. y(x, t) = 0.12 sin 2 x + 10t The phase is zero since sin(0) = 0 satisfies the indicated boundary condition. I suppose it is ok if the coefficient of the sin() function is 12 rather than 0.12, just remember that the units of the wave function are then cm rather than meters. d) (5 pts) Work out the maximum transverse speed of the string as the wave passes. Since the wave function of a transverse wave describes the transverse displacement of the medium as a function of both horizontal position and time, the transverse speed of the medium can be found by simply taking the rate of change of the transverse displacement (i.e. the wave function) with time: vy = @y @t vy = @ @t n0.12 sin 2 x + 10to vy = 1.2 cos 2 x + 10t The maximum transverse velocity occurs when the cos() term has its maximum value: 1. The maxi- mum transverse velocity is therefore 1.2 = 3.8 m/s. The other possiblilty, if you tried the approach of memorizing hundreds of special equations rather than being able to apply a few fundamental concepts, is to recall equation (13.15) from the text, vy,max = !A and just plug in the numbers.

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