Pendulum Problems

Examples using Huygen’s Law of for the period of a Pendulum. using the period, T of a pendulum depends on the square root of L, the length of the pendulum and g, the acceleration due to gravity.

Period of Pendulum

Additionally, the frequency f, and the period T, are reciprocals.    T = 1/f

Although we could use any unit for the period (years, months, eons, etc) the standard metric unit is the second.

The metric unit for frequency is Hertz which is one cycle per second.  However, another common units is rpm (rotations per minute).

Example 1:

What is the frequency of a pendulum with a length of 24″?

Givens: L = 24″     m per sec sq unit

Unknown: f = ???

First, convert the length into meters

L = 24″ =60.96 cm = 0.610 m

Next, plug the length and acceleration due to gravity into Huygen’s Law.

T = 2π √(.610/9.80) = 2π (0.2495) s = 1.57 s

Now that we have the period T, we take the reciprocal to find the frequency.

f = 1/T = 1/1.57 s = 0.647 Hz

Example 2:

What is the length of a pendulum with a period of 10 seconds?

Givens: T = 10.0 s m per sec sq unit

Unknown: L = ???

First, we take Huygen’s Law:


Period of Pendulum

and we algebraically isolate the variable L.  First we square both sides of the equation.

Squared Equation

Then cross-multiply to isolate L

isolate L

Then we plug in our values for T and g.

solve ex 2

Example 3:

Suppose you are an astronaut and need to find the acceleration due to gravity on an asteroid using a pendulum.  The period of the pendulum is 20 seconds and the length of the pendulum is 10 cm.  What is g?

Givens: L = 10 cm = 0.1 m       m per sec sq unit

Unknown: T = ???

First, we take Huygen’s Law:


Period of Pendulum

and we algebraically isolate the variable L.  First we square both sides of the equation.

Squared Equation

Then cross-multiply to isolate g

g equation

resulting in an acceleration of gravity of about

g = 0.01 m/s2