Transverse waves on a string - People.fas.harvard.edu

6 CHAPTER 4. TRANSVERSE WAVES ON A STRING
This equation says that for a given value of k, we can decompose the function C(k, t) into
its eiωt Fourier components. Plugging this expression for C(k, t) into Eq. (13) gives
∞ ∞
ψ(x, t) =
=
−∞
β(k, ω)e
−∞
iωt dω
e ikx dk
∞ ∞
β(k, ω)e i(kx+ωt) dk dω. (16)
−∞
−∞
This is a general result for any function of two variables; it has nothing to do with the wave
equation in Eq. (4). This result basically says that we can just take the Fourier transform
in each dimension separately.
Let’s now apply this general result to the problem at hand. That is, let’s plug the above
expression for ψ(x, t) into Eq. (4) and see what it tells us. We will find that ω must be
equal to ck. The function β(k, ω) is a constant as far as the t and x derivatives in Eq. (4)
are concerned, so we obtain
0 = ∂2ψ(x, t)
∂t2 ∂x2 ∞ ∞
=
β(k, ω)
−∞ −∞
∞ ∞
=
−∞
−∞
− c 2 ∂2 ψ(x, t)
2 i(kx+ωt) ∂ e
− c 2 ∂2ei(kx+ωt) ∂t 2
β(k, ω)e i(kx+ωt)
− ω 2 − c 2 (−k 2 )
∂x2
dk dω
dk dω. (17)
Since the e i(kx+ωt) exponentials here are linearly independent functions, the only way that
this sum can be zero for all x and t is if the coefficient of each separate e i(kx+ωt) term is
zero. That is,
β(k, ω)(ω 2 − c 2 k 2 ) = 0 (18)
for all values of k and ω. 2 There are two ways for this product to be zero. First, we can
have β(k, ω) = 0 for particular values of k and ω. This certainly works, but since β(k, ω)
indicates how much of ψ(x, t) is made up of e i(kx+ωt) for these particular values of k and ω,
the β(k, ω) = 0 statement tells us that this particular e i(kx+ωt) exponential doesn’t appear
in ψ(x, t). So we don’t care about how ω and k are related.
The other way for Eq. (18) to be zero is if ω 2 − c 2 k 2 = 0. That is, ω = ±ck, as we
wanted to show. We therefore see that if β(k, ω) is nonzero for particular values of k and ω
(that is, if e i(kx+ωt) appears in ψ(x, t)), then ω must be equal to ±ck, if we want the wave
equation to be satisfied.
4.2 Reflection and transmission
4.2.1 Applying the boundary conditions
Instead of an infinite uniform string, let’s now consider an infinite string with density µ1
for −∞ < x < 0 and µ2 for 0 < x < ∞. Although the density isn’t uniform, the tension
is still uniform throughout the entire string, because otherwise there would be a nonzero
horizontal acceleration somewhere.
Assume that a wave of the form ψi(x, t) = fi(x − v1t) (the “i” here is for “incident”)
starts off far to the left and heads rightward toward x = 0. It turns out that it will be much
2 Alternatively, Eq. (17) says that β(k, ω)(ω 2 − c 2 k 2 ) is the 2-D Fourier transform of zero. So it must be
zero, because it can be found from the 2-D inverse-transform relations analogous to Eq. (3.43), with a zero
appearing in the integrand.