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.

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″?

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?

Unknown: **L **= ???

First, we take Huygen’s Law:

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

Then cross-multiply to isolate **L**

Then we plug in our values for T and g.

# 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?

Unknown: T = ???

First, we take Huygen’s Law:

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

Then cross-multiply to isolate g

resulting in an acceleration of gravity of about

g = 0.01 m/s^{2}