Deriving an Expression for a Sinusoidal Wave on a String

[krazeekid7]

How would I get an expression y(x,t) that describes a sinusiodal wave traveling on a string in the negative x-direction with amplitude in the y-direction of 1.00cm, frequency 200Hz, and wavelength 3.00cm? At t=0, the particle of string at x=0 is displaced D=0.80cm from equilibrium and moving upwards.

 

[Tide]

Can you figure out what $\phi$ has to be?

$$y(x, t) = 1.00 \times \sin \left( 2 \pi \frac {x}{3.00} + 200 \times 2 \pi t + \phi\right)$$

 

[wais]

To get an expression for this scenario, we first need to understand the basic equation for a sinusoidal wave:

y(x,t) = A*sin(kx - ωt + φ)

where A is the amplitude, k is the wave number (2π/λ), ω is the angular frequency (2πf), and φ is the phase constant.

In this case, we are given A = 1.00cm, f = 200Hz, and λ = 3.00cm. Plugging these values into the equation, we get:

y(x,t) = 1.00cm*sin((2π/3.00cm)x - (2π*200Hz)t + φ)

Now, we need to determine the value of φ. We are given that at t=0, the particle at x=0 is displaced D=0.80cm from equilibrium and moving upwards. This means that the initial phase of the wave is such that the particle is at its maximum displacement (D = A = 1.00cm) and moving upwards (positive direction). This corresponds to a phase angle of π/2.

Substituting this value into the equation, we get our final expression for the wave:

y(x,t) = 1.00cm*sin((2π/3.00cm)x - (2π*200Hz)t + π/2)

This equation describes a sinusoidal wave traveling in the negative x-direction with an amplitude of 1.00cm and a frequency of 200Hz, with a wavelength of 3.00cm. At t=0, the particle at x=0 is at a displacement of 0.80cm from equilibrium and moving upwards.