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 = 14.0 cm, and the wave speed is v = 18.0 m/s. Furthermore, the wave is such that y = 0 at x = 0 and t = 0. Determine the angular frequency for this wave, rad/s Determine the wave number for this wave, rad/m Write an expression for the wave function. (Use the following as necessary: t, x. Let x be in meters and t be in seconds.) y = sin( ) Calculate the maximum transverse speed, m/s Calculate the maximum transverse acceleration of an element of the string.

Explanation / Answer

a) w = 2 pi f = 2* pi* 5 = 31.4
b) w/k = v

k = w/v = 31.4/18=1.74

c) y = A sin kx - wt

y = .14 sin ( 1.74 x - 31.4 t)

d) max tranvserse speed is max of dy/dt

= A w = .14*31.4=4.396

e) max acceleration is max of d^2y/dt^@

= A w^2 = .14*31.4^2 = 138.03

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