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 = 15.0 cm, and the wave speed is v = 22.0 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. m/s^2

Explanation / Answer

(a)
f = 5.0 Hz
Angular freq, w = 2**f
w = 2**5.0 rad/s
w = 31.42 rad/s

(b)
v = f *
= 22.0/5.0 = 4.4 m
Wave number, k = 2*/
k = (2*)/(4.4)
k = 1.43 rad/m

(c)
y = A * sin(kx - w*t)
y = 0.15 * sin(1.43 x - 31.42*t)

(d)
v = dy/dt
v = 0.15 * 31.42 m/s
v = 4.713 m/s

(e)
a = d^2y/dt^2
a = 0.15 * 31.42^2 m/s^2
a = 148.0 m/s^2

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