ROTATIONAL MOTION: The Frequency and Period of a Pendulum

Q: How may we derive an expression for the frequency ( f ) and period ( T ) of a pendulum? How does the motion of a rotating radius ( r ) within a unit circle relate to the centripetal or “ center seeking “ force that maintains a simple-pendulum system?

A: We begin by looking at a diagram of a simple pendulum in a vertical orientation:

If we slightly rotate the pendulum to the right, we have a system that contains a small angle relative to the vertical axis:

The angle in question must be small; otherwise, the physics of the system changes, and in addition to this, we will not be afforded the convenience of simplifying the system in a manner that is desirable. The pendulum swings back and forth around a central region, and for this reason, the force that drives it to-and-fro is a centripetal force ( F_c ). Additionally, the system is driven by a component of the gravitational force ( F_g ) of attraction acting downward. If the angle of displacement is small enough, we use the ( sin θ ) function to represent the component of force that acts perpendicular to the arm of the pendulum:

F_c =  F_g sin θ

F_g = mg

F_c = mg sin θ

Note: Our derivations should have negative values, because the restoring force opposes the direction of travel of the pendulum when it swings past the origin; however, the negative signs eventually cancel out.

In addition to this, the centripetal force acting upon the system can be represented using the more-familiar expression for centripetal force:

F_c = mv² / r

The radius ( r ) of the system is the length ( L ) of the suspension in question. Setting these expressions equal to one another yields the following equation:

mv² / r = mg sin θ

v² / r = g sin θ

This is the equation for centripetal acceleration. With a relatively small angle, the sin θ function can be approximated using the following ratio:

sin θ = ( opp / hyp )

sin θ = ( l / r )

In the lattermost expression, ( l ) regards the relatively short distance through which the pendulum has been displaced from its origin. In addition to systems that contain a pendulum, a unit circle can be used to prove that the ratio of ( l ) divided by ( r ) is how an angle θ is approximated:

Regardless of the length of a circle’s radius, an angle θ will be represented by a ratio of a subtended distance along its outer surface and the radius at hand. The longer the radius is, the longer the surface area subtended by an angle, but for any given angle, this ratio will be the same. And, for these reasons…

v² / r = g sin θ

v² / r = ( g )( l / r )

v² / r = ( g / r )( l )

Recall that the radius of the pendular system is equal to ( L ), and for this reason, another substitution is appropriate:

v² / r = ( g / L )( l )

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The second-half of our derivation involves an algebraic manipulation of θ:

θ = ( l / r ) 

And thus,

l = rθ

If the unit circle’s radius were to rotate at a fixed rate, the distance along the outer circle would acquire a velocity as follows:

velocity = ( meters / second ) = ( l / s )

Of course, we must divide the right hand side of the equation by seconds as well. If we divide the θ term by seconds, we get an expression for angular speed ( ⍵ ):

⍵ = ( θ / s )

And thus,

( l / s ) = ( r )( θ / s )

v = r⍵

v² = r²⍵²

v² / r = r⍵²

We now have two expressions for centripetal acceleration, and thus we are at liberty to set them equal to one another:

( g / L )( l ) = r⍵²

( g / L )( l / r ) = ⍵²

In the beginning stages of our derivations using a pendulum, we saw the following:

sin θ = ( l / r )

And thus,

( g / L )( sin θ ) = ⍵²

Note: The initial condition was that the angle θ be small. For this reason, the sin θ part of the equation is a small and unitless quantity that can be omitted at this point in our journey:

( g / L ) = ⍵²

⍵ = √ ( g / L )

Another useful expression can be derived by multiplying angular speed by the period ( T ) of time needed for a complete revolution of 2π radians to transpire:

⍵T = 2π

⍵ = ( 2π / T )

Furthermore, the frequency ( f ) of a rotating system is the inverse of its period:

f = ( 1 / T )

And thus,

⍵ = 2πf

And alas, we have our expressions for the period and frequency of a pendulum:

2πf = √ ( g / L )

f = 1 / 2π √ g / L

And,

( 1 / T ) = 1 / 2π √ g / L

T = 2π √ L / g