Simple Nature - Light and Matter

n/A triangular pulse spreads out.

As always, the velocity of a wave depends on the properties of

the medium, in this case the string. The properties of the string can

be summarized by two variables: the tension,T, and the mass per

unit length,μ(Greek letter mu).

If we consider the part of the string encompassed by the initial

dent as a single object, then this object has a mass of approxi-

matelyμw(mass/length×length=mass). (Here, and throughout

the derivation, we assume thathis much less thanw, so that we can

ignore the fact that this segment of the string has a length slightly

greater thanw.) Although the downward acceleration of this seg-

ment of the string will be neither constant over time nor uniform

across the pulse, we will pretend that it is constant for the sake of

our simple estimate. Roughly speaking, the time interval between

n/1 and n/2 is the amount of time required for the initial dent to ac-

celerate from rest and reach its normal, flattened position. Of course

the tip of the triangle has a longer distance to travel than the edges,

but again we ignore the complications and simply assume that the

segment as a whole must travel a distanceh. Indeed, it might seem

surprising that the triangle would so neatly spring back to a per-

fectly flat shape. It is an experimental fact that it does, but our

analysis is too crude to address such details.

The string is kinked, i.e., tightly curved, at the edges of the

triangle, so it is here that there will be large forces that do not

cancel out to zero. There are two forces acting on the triangular

hump, one of magnitudeT acting down and to the right, and one

of the same magnitude acting down and to the left. If the angle

of the sloping sides isθ, then the total force on the segment equals

2 Tsinθ. Dividing the triangle into two right triangles, we see that

sinθequalshdivided by the length of one of the sloping sides. Since

his much less thanw, the length of the sloping side is essentially

the same asw/2, so we have sinθ= 2h/w, andF = 4Th/w. The

acceleration of the segment (actually the acceleration of its center

of mass) is

a=

F

m

=

4 Th

μw^2

.

The time required to move a distancehunder constant acceleration
ais found by solvingh= (1/2)at^2 to yield

t=

√

2 h/a

=w

√

μ

2 T

.

Section 6.1 Free waves 361