E Classical Mechanics 1269
String
xx
xx 1 ∆x
z
F 1
F 2
1
2
Figure E.1 The Position of a Flexible String and the Forces on a Segment of the
String.This arbitrary conformation of the string as a function of length along the string is
used to describe the forces on a small segment of the string.
α 1 andα 2. For small displacements, the net force on the string segment will lie in the
zdirection:
FzF 2 z+F 1 zTsin(α 2 )−Tsin(α 1 )≈T[tan(α 2 )−tan(α 1 )]
≈T
[(
∂z
∂x
)∣
∣
∣
∣
x′+∆x
−
(
∂z
∂x
)∣
∣
∣
∣
x′
]
(E-7)
where the subscripts on the derivatives denote the positions at which they are evaluated.
We have used the fact that the sine and the tangent are nearly equal for small angles,
which corresponds to our case of small displacements. We have also used the fact that
the first derivative is equal to the tangent of the angle between the horizontal and the
tangent line.
The mass of the string per unit length isρand the mass of the segment isρ∆x. Its
acceleration is (∂^2 z/∂t^2 ), so that from Newton’s second law
T
[(
∂z
∂x
)∣∣
∣
∣
x+∆x
−
(
∂z
∂x
)∣∣
∣
∣
x
]
ρ∆x
∂^2 z
∂t^2
(E-8)
We divide both sides of this equation byρ∆xand then take the limit as∆xis made to
approach zero. In this limit, the quotient of differences becomes a second derivative:
lim
∆x→ 0
⎡
⎢
⎢
⎢
⎣
(
∂z
∂x
)∣∣
∣
∣
x+∆x
−
(
∂z
∂x
)∣∣
∣
∣
x
∆x
⎤
⎥
⎥
⎥
⎦
∂^2 z
∂x^2
(E-9)
This equation is substituted into Eq. (E-8) to obtain the classical wave equation:
∂^2 z
∂x^2
1
c^2
∂^2 z
∂t^2
(E-10)
wherec^2 T/ρ. We show in Chapter 14 thatcis the speed of propagation of a traveling
wave in the string.