The string shown in the figure is driven at a frequency of 5.00 Hz. The amplitud

Question

The string shown in the figure is driven at a frequency of 5.00 Hz. The amplitude of the motion is A = 12.5 cm, and the wave speed is v = 18.5 m/s. Furthermore, the wave is such that y = 0 at x = 0 and t = 0. Determine the angular frequency for this wave, Determine the wave number for this wave, Write an expression for the wave function. (Use the following as necessary: t, x. Let x be in meters and t be in seconds.) Calculate the maximum transverse speed, Calculate the maximum transverse acceleration of an element of the string.

Explanation / Answer

a) w = 2 pi f = 2*pi*5 = 31.4 rad/s

b) v = w/k

k = w/v = 31.4/18.5= 1.7

c)
y = A sin( kx - wt)
y = .125 sin( 1.7 x - 31.4 t)

d) max is A w = .125*31.4= 3.93

e) max accel is Aw^2 = .125*31.4^2 =123

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