The initial phase angle of the transverse wave is ∅ = 0.3787 rad
<u>Given that</u>
period T= 25.0ms
speed of 30.0 m/s
t = 0
x = 0
transverse position of 2.00cm
traveling downward with a speed of 2.0 m/s
<h3>how to solve for the initial phase angle</h3>
transverse position of 2.00cm = 0.02m
wave equation is = y ( x ) = A cos ( ωx + ∅ )
since:
t = 0
x = 0
0.02 = A cos ( 0 + ∅ )
0.02 = A cos ( ∅ )
speed = v ( x )= 2.0 m/s = d y ( x ) / d x
- 2 = - Aω sin ( ∅ )
dividing the equations as below gives
- Aω sin ( ∅ ) / A cos ( ∅ )
- ω tan ( ∅ ) = - 2 / 0.02
ω tan ( ∅ ) = 100
ω = 2 * pi * f = 2 * pi / T = 2 * pi / 0.025 = 80 * pi
tan ( ∅ ) = 100 /( 80* pi )
tan ( ∅ ) = 0.398
∅ = 0.3787 rad
Read more on phase angle here: brainly.com/question/7956945
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