Differential equations; wave equation

[briteliner]

Homework Statement

the transverse displacement y(x,t) (assumed small) of a string stretched between the points x=0 and x=a satisfies the equation
d²y/c²dt²=d²y/dx²

find the solution satisfying the following (t=0) initial conditions
(i) y(x,0)= Lsin($$\pi$$x/a), (dy/dt)(x,0)=0
all derivatives are partial.

Homework Equations

The Attempt at a Solution

there are more parts to the problem but if i can get started on this one i think i can figure them out. i don't know where to start since the y(x,0) doesn't depend on t. so when i take the partial with respect to t would it just be 0??

 

[mighty2000]

Thank you for your question. To solve this problem, we will first need to understand the physical meaning of the equation and initial conditions provided.

The equation d2y/c2dt2=d2y/dx2 is known as the wave equation and it describes the motion of a string under tension. In this equation, c represents the speed of the wave, which is related to the tension and density of the string. The variable y(x,t) represents the displacement of the string at position x and time t.

Now, let's look at the initial conditions provided. The first condition, y(x,0)= Lsin(\pix/a), tells us the initial displacement of the string at time t=0. This means that when time t=0, the string is in the shape of a sine wave with amplitude L and wavelength 2a. The second condition, (dy/dt)(x,0)=0, tells us that the initial velocity of the string is 0 at all points x. This means that the string is not moving at all at time t=0.

To solve this problem, we will need to use the method of separation of variables. We will assume that the solution can be written as a product of two functions, one depending only on x and the other depending only on t. So, y(x,t)=X(x)T(t). Substituting this into the wave equation, we get:

X''(x)T(t)/c^2 = X(x)T''(t)

We can rearrange this equation to get:

X''(x)/X(x) = T''(t)/c^2T(t)

Since the left side of the equation only depends on x and the right side only depends on t, both sides must be equal to a constant, which we will call -k^2. So, we now have two separate equations:

X''(x) + k^2X(x) = 0

T''(t) + c^2k^2T(t) = 0

The first equation is known as the spatial equation and the second equation is known as the temporal equation.

Solving the spatial equation, we get X(x)=Asin(kx)+Bcos(kx). Applying the first initial condition, y(x,0)= Lsin(\pix/a), we get:

Lsin(\pix/a) = Asin(0