Calculating Coordinate of B in Sinusoidal Wave Equation

[dekoi]

At a certain instant, a point A is at the origin and a point B is the first point on the x-axis that is 60° out of phase. What is the coordinate of B?

The equation of the wave is given as:
y = (15.0 cm)cos(0.157x-50.3t)

 

[SpaceTiger]

At a certain instant, a point A is at the origin and a point B is the first point on the x-axis that is 60° out of phase. What is the coordinate of B?

The equation of the wave is given as:
y = (15.0 cm)cos(0.157x-50.3t)

Hint:

$$y = A cos(kx+\omega t)$$

where A is the amplitude of the wave, k is the wavenumber, and $\omega$ is the angular frequency. The latter two are given by

$$k=\frac{2\pi}{\lambda}$$
$$\omega=\frac{2\pi}{T}$$

where T and $\lambda$ are the period and wavelength. What fraction of a wavelength is between the two points?

 

[quicknote]

To calculate the coordinate of point B, we need to first determine the value of x at 60° out of phase. This can be done by using the formula:
phase angle (θ) = (x/λ) * 2π

Where,
x = distance traveled by the wave
λ = wavelength

In this case, we know that the wavelength (λ) is equal to the distance between point A and B on the x-axis, which is equal to the amplitude (15.0 cm) of the wave.

Therefore, we can rearrange the formula to solve for x:
x = (θ * λ) / 2π

Substituting the given values, we get:
x = (60° * 15.0 cm) / 2π = 4.774 cm

Now, to find the coordinate of point B, we can simply plug in the value of x into the equation for the wave:
y = (15.0 cm)cos(0.157(4.774 cm)-50.3t)

At a certain instant, t, we can solve for the coordinate of point B by substituting the value of t into the equation.

Thus, the coordinate of point B is (4.774 cm, 0). This means that at 60° out of phase, point B is located at a distance of 4.774 cm from the origin on the x-axis and has a y-coordinate of 0.