Answer:
The answer is below
Explanation:
Given that:
y = transverse displacement = 4.2 cm = 0.042 m at x = 0 and t = 0.
Speed = v = 89 m/s, maximum transverse speed of the string particle = [tex]u_m[/tex] = 16 m/s.
ω = [tex]u_m[/tex] / [tex]y_m[/tex] = 16 / 0.42 = 380.95 rad/s
a) Frequency = ω/2π = 380.95 / 2π = 60.63 Hz
b) Wavelength (λ) = speed / frequency
λ = v / f = 89/63.66= 1.468 m
c) Using the wave equation:
[tex]y=y_msin(kx \pm wt \pm \phi)\\y=0.042,t=0,x=0\\\\Hence\\y_m=0.042\ m[/tex]
d) Wave number k is given by:
k = 2π / λ = 2π / 1.468 = 4.28 rad/s
e) The angular velocity is given by:
ω = [tex]u_m[/tex] / [tex]y_m[/tex] = 16 / 0.42 = 380.95 rad/s
f) Using the wave equation:
[tex]y=y_msin(kx \pm wt \pm \phi)\\\\y=0.042,t=0,x=0,y_m=0.042\\\\Hence\\0.042=0.042sin(4.28(0)\pm 380.95(0)\pm \phi)\\\\sin\phi=1\\\\\phi=\frac{\pi}{2} \\\\y=0.042sin(4.28x\pm 380.95t\pm \frac{\pi}{2})[/tex]
g) Since the wave is in the positive x direction, hence ω is negative
