216 Chapter 3 The Wave Equation
Figure 1 String fixed at the ends.Figure 2 Section of string showing forces exerted on it. The angles are
α=φ(x,t)andβ=φ(x+ x,t).
or
T(x,t)cos(
φ(x,t))
=T(x+ x,t)cos(
φ(x+ x,t))
. (1)
This says that the horizontal component of tension in the string is the same at
every point:
T(x,t)cos(
φ(x,t))
=T(x+ x,t)cos(
φ(x+ x,t))
=T,
independent ofx. If the string is taut,Tcan vary only slightly witht,sowewill
assume thatTis constant.
In the absence of external forces other than gravity, Newton’s second law for
the vertical direction yields
−T(x,t)sin(
φ(x,t))
+T(x+ x,t)sin(
φ(x+ x,t))
−mg=m∂^2 u
∂t^2 (x,t).
(2)(Becauseu(x,t)measures displacement in the vertical direction, its second
partial derivative with respect totis the vertical acceleration.) The mass of the
short piece of string we are examining is proportional to its length,m=ρ x,
whereρis the linear density, measured in units of mass per unit length.
Now we use Eq. (1) to solve for the tensions at the ends of the segment of
string as
T(x,t)= T
cos(φ(x,t)), T(x+ x,t)= T
cos(φ(x+ x,t))