13 WAVE MOTION 13.2 Waves on a stretched string

Here, ∂^2 y(x, t)/∂x^2 is the second derivative of y(x, t) with respect to x, keeping t

constant.

Suppose that the string has a mass per unit length μ. It follows that the y

equation of motion of our string segment takes the form

∂^2 y(x, t)

μ δx = fy(x, t), (13.5)

∂t^2

Here, ∂^2 y(x, t)/∂t^2 —the second derivative of y(x, t) with respect to t, keeping

x constant—is the y-acceleration of the string segment at position x and time t.

Equations (13.4) and (13.5) yield the final expression for the string’s equation of

motion:

∂^2 y

∂t^2

T ∂^2 y

=

μ ∂x^2

. (13.6)

Equation (13.6) is an example of a wave equation. In fact, all small amplitude

waves satisfy an equation of motion of this basic form. A particular solution of

this type of equation has been known for centuries: i.e.,

y(x, t) = y 0 cos (k x − ω t), (13.7)

where y 0 , k, and ω are constants. We can demonstrate that (13.7) satisfies (13.6)

by direct substitution. Thus,

∂y

(^)
∂t
∂^2 y
(^)
= y 0 ω sin (k x − ω t), (13.8)
(^2)
and
∂t^2
= −y 0 ω
∂y
cos (k x − ω t), (13.9)
(^)
∂x
∂^2 y
(^)
= −y 0 k sin (k x − ω t), (13.10)
(^2)
∂x^2
= −y 0 k cos^ (k^ x^ −^ ω^ t).^ (13.11)^
Substituting Eqs. (13.9) and (13.11) into Eq. (13.6), we find that the latter equa-
tion is satisfied provided
ω^2 T

k^2 μ

. (13.12)