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

4.3. IMPEDANCE 13
In the case of uniform tension discussed in the previous section, we had T1 = T2, so this
equation reduced to the first equality in Eq. (24). With the tensions now distinct, the only
modification to the second equality in Eq. (24) is the extra factors of T1 and T2. So in terms
of the f functions, Eq. (34) becomes
− T1
v1
f ′ i (t) + T1
v1
f ′ r(t) = − T2
f ′ t(t). (35)
The other boundary condition (the continuity of the string) is unchanged, so all of the
results in the previous section can be carried over, with the only modification being that
wherever we had a v1, we now have v1/T1. And likewise for v2. The quantity v/T can be
written as
where
v
T =
T/µ
T
v2
= 1
√ ≡
T µ 1
, (36)
Z
Z ≡ T
v = T µ (37)
is called the impedance. We’ll discuss Z in depth below, but for now we’ll simply note that
the results in the previous section are modified by replacing v1 with 1/ √ T1µ1 ≡ 1/Z1, and
likewise for v2. The reflection and transmission coefficients in Eq. (31) therefore become
R =
1
Z2
1
Z2
− 1
Z1
+ 1
Z1
= Z1 − Z2
Z1 + Z2
Note that Z grows with both T and µ.
Physical meaning of impedance
and T =
1
Z2
2
Z2
+ 1
Z1
=
2Z1
Z1 + Z2
What is the meaning of the impedance? It makes our formulas look nice, but does it have
any actual physical significance? Indeed it does. Consider the transverse force that the ring
applies to the string on its left. Since there is zero net force on the ring, this force also
equals the transverse force that the right string applies to the ring, which is
∂ψR(x, t)
Fy = T2
∂x
x=0
∂ft(t − x/v2)
= T2
∂x
(38)
, (39)
x=0
where we have labeled the transverse direction as the y direction. But the chain rule tells
us that the x and t partial derivatives of ft are related by
∂ft(t − x/v2)
∂x
= − 1
v2
· ∂ft(t − x/v2)
∂t
. (40)
Substituting this into Eq. (39) and switching back to the ψR(x, t) notation gives
Fy = − T2
·
v2
∂ψR(x,
t)
∂t = −
x=0
T2
· vy ≡ −bvy, (41)
v2
where vy = ∂ψR(x, t)/∂t is the transverse velocity of the ring (at x = 0), and where b is
defined to be T2/v2.
This force Fy (which again, is the force that the ring applies to the string on its left) has
the interesting property that it is proportional to the (negative of the) transverse velocity.
It therefore acts exactly like a damping force. If we removed the right string and replaced
the ring with a massless plate immersed in a fluid (in other words, a piston), and if we