Simple Nature - Light and Matter

e/Water in a U-shaped tube.

so the energy in the spring is

U=

1

2

k

(

L

b

x

) 2

=

k L^2

2 b^2

x^2 ,

and thekwe have to put inT= 2π

√

m/kis

d^2 U

dx^2

=

k L^2

b^2

.

The result is

T= 2π

√

mb^2

k L^2

=

2 πb

L

√

m

k

The leverage of the lever makes it as if the spring was stronger,

and decreases the period of the oscillations by a factor ofb/L.

Water in a U-shaped tube example 24

.What is the period of oscillation of the water in figure e?

.In example 13 on p. 89, we foundU(y) =ρgAy^2 , so the “spring

constant,” which really isn’t a spring constant here at all, is

k=

d^2 U

dy^2

= 2ρgA.

This is an interesting example, becausekcan be calculated with-

out any approximations, but the kinetic energy requires an ap-

proximation, because we don’t know the details of the pattern

of flow of the water. It could be very complicated. There will

be a tendency for the water near the walls to flow more slowly

due to friction, and there may also be swirling, turbulent motion.

However, if we make the approximation that all the water moves

with the same velocity as the surface, dy/dt, then the mass-on-

a-spring analysis applies. LettingLbe the total length of the filled

part of the tube, the mass isρLA, and we have

T= 2π

√

m/k

= 2π

√

ρLA

2 ρgA

= 2π

√

L

2 g

.

This chapter is summarized on page 1072. Notation and terminology
are tabulated on pages 1066-1067.

Section 2.5 Oscillations 119