Wave Motion of sinusoidal wave

[magisbladius]

Homework Statement

For a sinusoidal wave:

At a certain instant, let point A be at the origin and point B be the first point along the x-axis where the wave is 60.0° out of phase with point A. What is the coordinate of point B?

y = (15.0 cm) cos(0.157x - 50.3t)

Homework Equations

The Attempt at a Solution

x(at point B) = theta + 0 = 60.0° = pi/3 rad = ~1.05 m.
y(at point B) = (0.150 m) cos(0.157[1.047 m] - 50.3[0]) = 0.148 m

(x, y) = (1.05, 0.148) m

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I don't understand what the question is asking for. I am handing this in, online, and it checks the answers (I have 3 tries left).

 

[magisbladius]

I found the answer.

 

[Moetasim]

Based on the given information, point B is located at a position where the wave is 60.0° out of phase with point A, which is at the origin (x=0). The equation provided represents the wave motion of a sinusoidal wave, where y is the displacement of the wave at a certain point, x is the distance along the x-axis, and t is the time.

To find the coordinate of point B, we need to substitute the given values into the equation. We know that the phase difference between point A and B is 60.0° or pi/3 radians, which is represented by the term (0.157x - 50.3t). Therefore, we can write the equation as:

y = (15.0 cm) cos(0.157x - 50.3t) = (15.0 cm) cos(pi/3)

Solving for the cosine of pi/3, we get a value of 0.5. Substituting this back into the equation, we get:

y = (15.0 cm) (0.5) = 7.5 cm

Therefore, the coordinate of point B is (x,y) = (1.05 m, 7.5 cm).

Note: The given equation is in terms of meters, while the given values for amplitude and wavelength are in centimeters. This is why we have converted the amplitude to meters in our solution.