Boundary Conditions for Wave Solution

Classical Mechanics Level 2

Given an arbitrary harmonic solution to the wave equation

$y(x,t) = A \sin (x-vt) + B \sin (x+vt),$

find the general solution, i.e. find the coefficients $A$ and $B$ given the following boundary conditions:

$y(0,t) = 0, \qquad y(L,0) = 1.$

y(x,t) =\frac{\sin x}{\sin L}

y(x,t) =\sin x \cos L \cos vt

y(x,t) =\sin x \sin (L-vt)

y(x,t) = \frac{\sin x \cos vt}{\sin L}

1 solution

Matt DeCross
May 10, 2016

The condition $y(0,t) = 0$ gives the condition $B-A = 0$ on the coefficients, i.e. $A=B$ . The condition $y(L,0) = 1$ then gives:

$1 = A \sin L + A \sin L \implies A = \frac{1}{2\sin L}.$

Substituting in, one finds the solution:

$y(x,t) = \frac{1}{2\sin L} \sin (x-vt) + \frac{1}{2\sin L} \sin (x+vt) = \frac{\sin x \cos vt}{\sin L}.$

How was the cos gotten

Anthony Onyia - 1 year, 2 months ago

If you apply the angle sum/difference formula for sine:

sin(A+B) = sinAcosB+cosAsinB, sin(A-B) = sinAcosB - cosAsinB

you'll get the cosine term in the numerator above.

tom engelsman - 7 months, 1 week ago

product to sum formula : sin a * cos b = 1/2 [ sin (a + b) + sin ( a - b) ]

Austin Chang - 4 months ago

That is what I put and it was “incorrect “

hunt Bobo - 2 months ago