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Worden Zero-length Spring Pendulum Analysis

B. H. Suits, Physics Dept, Michigan Tech University, 2005.

See the figures for geometry and definition of variables.

The torque due to gravity (clockwise) is given by

. g Mga= cos

The torque due to the spring (counter-clockwise) is given by

.

s s sF Fbd

l bkd

l l

l= = =

sincos ( )

cos0

The equilibrium spring length, leq, is determined by the condition for zero

torque

.bkdl l

l Mga

eq

eq

=0

(However, if the ls cancel and one simply has the conditionl0 0=

k Mga bd = /

which has no or ldependence, corresponding to a very flatequilibrium or none at all that is, even a minor change in the force due

to gravity can make a huge change in the position.)

So, assuming l00,

.l lbkd

bkd Mgal

Mga

bkdeq =

=

0 0

1

1

The net torque can be written

=

=

Mga bkd

l l

lbkl d

l leq0

0

1 1cos cos

whereMgawas replaced using the previous result. The changes in the torque with angle about equilibrium are given

by

d

dbkl d

l

dl

d l leq eq

eq

=

0 2

1 1 1sin

and

dl

d

bd

b d bd

bd

leq eq

eq

eq

=

+ +=

cos

( sin )

cos

/2 2 1 22

and so if we now consider only the case where the system is adjusted so equilibrium occurs when = 0,

.d

dkl bd

b deq

=

+0

2

2 2 3 2

( )

( ) /

Near equilibrium, the equation of motion gives

+

=klbd

b dMa

0

2

2 2 3 2

2( )

( )( )&&

/

whereMa2is the moment of inertia for rotations assuming the mass of the support structure and spring are negligible

and thatMcan be treated as a point mass. If that is not the case, a more correct moment of inertia should be used.

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This can be written

&&

( ) /

/

= +

=

bd

a

kl

M b d

0

2 2 3 2

1 2 2

2

so the solutions are sinusoidal in time with angular frequency

. =+

=bd

akl

M b dbd

alk

Mlleq eq

0

2 2 3 2

1 2

0

( ) /

/

This particular result for the frequency is only for the case where the equilibrium position has been adjusted to be at

= 0, that is , and not the most general case. See the above equations if you wish to derive thel b deq = +2 2

general case. It is possible to use other equilibrium positions with equivalent results.

Note that when l060, the frequency goes to zero (the period becomes infinite). That situation is analogous to

placing an object on a frictionless table. If the table is perfectly horizontal (in this case, the parameters a, b, d, k,M

are just right, that is kbd/Ma= g) then the object will stay wherever you put it. If any of these parameters (or g) is

changed, the object rolls toward infinity until it falls off the table. In practice, finding a spring with l0exactly zero is

impossible and real springs do not obey Hookes law and the response will be finite.

When used to measure small changes in g, you can estimate the sensitivity by looking only at the equilibrium

position. In actual instruments, what one usually determines is how much a given parameter must be changed to

bring the system back to equilibrium if there is a change in g.