Progressive wave with initial conditions f(x,0)=0, f'(x,0)=g(x)

[wakko101]

Consider a long straight string that is given an initial impulse. The transverse displacement of the string y(x,t) satisfies the initial condition:

y(x,0) = 0 and y'(x,0) = G(x)

Show that the solution to the wave eq'n satisfying the intitial condition is

y(x,t) = 1/2v[H(x+vt)-H(x-vt)]

where H'(u) = G(u)

Solve for an impulse about x=0 with a step profile of the form:

0 x less than/equal to -a
-b -a < x less than/equal to 0
b 0 < x less than/equal to a
0 x > a

where a and b are positive constants.

I'm not entirely sure where to start with this. In class, we derived a solution for initial conditions of f(x,0) = f(x) and f'(x,0) = 0 by using the standing wave eq'n. Physically, that represented a long string with an initial displacement, released from rest. It's easier to picture that physical situation. This is the opposite, whereby the string has no initial displacement, but an initial velocity. Should I start with the standing wave eq'n again? Or should I go back to the wave equation (d^2f/dx^2=v^2(d^2f/dt^2)?

Any suggestions or clarifications would be appreciated.

Cheers,
W. =)

 

[Jhero]

Dear W.,

Thank you for your post. It seems like you are on the right track with using the wave equation to solve this problem. Let's start with the general solution to the wave equation:

y(x,t) = f(x-vt) + g(x+vt)

where f and g are arbitrary functions. This solution represents a wave traveling in the positive x direction with velocity v. Now, let's plug in the initial conditions given in the problem:

y(x,0) = 0 and y'(x,0) = G(x)

This gives us the following equations:

f(x) + g(x) = 0
-vf'(x) + vg'(x) = G(x)

From the first equation, we can solve for g(x) as g(x) = -f(x). Substituting this into the second equation, we get:

-vf'(x) - vf'(x) = G(x)
-2vf'(x) = G(x)
f'(x) = -G(x)/(2v)

Now, we can integrate both sides to solve for f(x):

f(x) = -1/2v * ∫G(x) dx

Substituting this back into our solution for y(x,t), we get:

y(x,t) = -1/2v * ∫G(x-vt) dx + g(x+vt)

Now, let's use the fact that H'(u) = G(u). This means that we can rewrite the integral as:

y(x,t) = -1/2v * ∫H(x-vt) dx + g(x+vt)

Now, we need to find the function g(x+vt) that satisfies the initial condition y'(x,0) = G(x). We can do this by taking the derivative of g(x+vt) and setting it equal to G(x):

g'(x+vt) = G(x)

Integrating both sides with respect to x, we get:

g(x+vt) = ∫G(x) dx + C

where C is a constant of integration. Finally, we can substitute this back into our solution for y(x,t):

y(x,t) = -1/2v * ∫H(x-vt) dx + ∫G(x) dx + C

Now, let's use the fact that H(u) is a step function, which means that it is equal to