Calculating Phase Angle and Amplitude for a Sinusoidal Wave on a String

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SUMMARY

The discussion focuses on calculating the phase angle and amplitude of a rightward traveling sinusoidal wave on a string with a frequency of 230 Hz. The displacement at time t = 0 and position x = 0 m is given as y = +4.8 mm, and the transverse velocity is vy = -0.75 m/s. The smallest positive phase angle is determined to be approximately 1.678 radians, while the amplitude of the wave is calculated to be 0.00483 m. The time required for an observer at x = 0 to wait before a trough arrives is calculated using the formula for angular frequency.

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Homework Statement



At time t = 0 and at position x = 0 m along a string, a rightward (+x dir) traveling sinusoidal wave with frequency of 230 Hz has displacement y = +4.8 mm and transverse velocity vy = -0.75 m/s.

What is the smallest positive phase angle?

What is the amplitude of the wave?

How long does an observer at x=0 need to wait before a trough (y=-A) arrives?





The Attempt at a Solution



.0048=Asin(phi)
-.75=acos(phi)

arctan[2(pi)(230)(.0048/-.75)]=phi=-1.463

pi-1.463=1.678rad

For some reason this answer is incorrect and I don't know why since I used in my calculation of part B which is correct.

B) .0048/sin(1.678)=.00483m

Not sure how to approach part c...

As always thanks for the help
 
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for c I would just solve t for -A correct, omega would be negative since the wave is moving rightward?

arcsin(-1)/(-2pi*230)=t=.0011

Does this make sense?
 

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