Download Derivation of the One-Dimensional Wave Equation (Vibrating String)

Derivation of the One-Dimensional Wave Equation (Vibrating String)
Consider a string which vibrates under the action of internal tension and at equilibrium, lies along the
x-axis.
Assumptions:
1. When the string vibrates, the motion takes place in the x-u-plane (i.e. the motion is planar) and
each point on the string moves in a direction perpendicular to the x-axis (i.e., we have transverse
vibrations).
2. The string is perfectly flexible (or elastic).
3. The tension acts tangentially to the string at each point.
4. The tension is large compared to the force of gravity and no other external forces act on the string.
Two implications of these assumptions are as follows:
1’. From 1, the net horizontal force on any section of the string is zero.
2’. From 2, there is no resistance to bending, thus, the shape of the string does not produce any
non-tangential force on the string.
Let u (x, t) be the displacement from the x-axis at time t.
Consider a segment of the string from a point x to a point x + h, h > 0:
Enlarge this section and consider t as fixed (“temporarily”):
Note: T (x) is the magnitude of the tension at x (and similarly for T (x + h) ).
T (x) and θ (x) are actually functions of x and t, but we drop the explicit dependence on t.
Derivation of the One-Dimensional Wave Equation - Page 2
From assumption 1’, the net horizontal force on the section [x, x + h] is
T (x + h) cos[θ (x + h)] − T (x) cos[θ (x)] = 0 .
Thus,
T (x + h) cos[θ (x + h)] = T (x) cos[θ (x)] = H
(1)
where H is a constant, since x and h are arbitrary points on the string.
Let ρ (x) be the linear density (mass per unit length) of the string at equilibrium.
The mass, m, of the section [x, x + h] is the same as its rest mass which is equal to
x+h
m=
∫
ρ (x)d x = h ρ (x + h1 ) , for some 0 < h1 < h , using the Mean-Value Theorem.
x
We apply Newton’s second law in the vertical direction:
net force equals mass × (average) acceleration.
T (x + h) s i n[θ (x + h)] − T (x) s i n[θ (x)] = h ρ (x + h1 ) u t t (x + h 2 , t)
(2)
for some 0 < h 2 < h .
We divide equation (2) by the expressions in (1) (which are equal):
h ρ (x + h1 ) u tt (x + h 2 , t)
T (x + h) s i n[θ (x + h)] T (x) s i n[θ (x)]
−
=
,
T (x + h) cos [θ (x + h)] T (x) cos [θ (x)]
H
giving
tan [θ (x + h)] − tan [θ (x)] =
h ρ (x + h1 ) u tt (x + h 2 , t)
.
H
From assumptions 2’ and 3, the only force on the string is tension which acts in a direction that is
tangent to the string. Hence, tan [θ (x)] = u x (x, t) , for every x, and this gives:
u x (x + h, t) − u x (x, t) =
Divide by h:
h ρ (x + h1 ) u tt (x + h 2 , t)
.
H
ρ (x + h1 ) u tt (x + h 2 , t)
u x (x + h, t) − u x (x, t)
=
.
h
H
Let h → 0 :
u xx (x, t) =
Let ρ (x) = ρ , a constant and let c =
ρ (x) u tt (x, t)
.
H
H
, which has the units of speed. Then, the above partial
ρ
differential equation becomes:
c 2 u xx = u tt
and this is our one-dimensional wave equation for the vibrating string.