solve The string shown in the figure in the figure is driven at a frequency of 5

Question

solve

The string shown in the figure in the figure is driven at a frequency of 5.00 Hz. The amplitude of the motion is A=14.5cm, and the wave speed is v=20.5m/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. Calculated the maximum transverse acceleration of an element of the string.

Explanation / Answer

y = Asin[2(x - vt)/] is the wave equation = Asin(kx - t )

(a) angular frequency = 2f=2*5=31.4 rad/s

(b) Wave number k =/v=31.4/20.5=1.53 rad/m

(c) y = Asin(kx - t) = 0.145sin(1.53x31.4t)

(d)  transverse speed is dy/dt = 0.145cos(1.53x31.4t) [-31.4]

or dy/dt is maximum when kx - t =

substituting this is the original equation => dy/dt (max) = 0.145(31.4) = A = 4.553 m/s

e) transverse acceleration = d²y / dt² = 4.553sin(1.53x31.4t)[-31.4]

this is maximum when (1.53x31.4t) = -½

and maximum acceleration is = A() = A² = 142.96 m/s²

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