# Show that the oscillation of a pendulum is a simple harmonic?

##### 1 Answer
Apr 4, 2017

This is the solution we get at the end:

$\vec{\theta} \left(t\right) = {\vec{\theta}}_{0} \cos \left(\sqrt{\frac{\vec{g}}{L}} t\right)$ $\text{ }$$\boldsymbol{\left(6\right)}$

$T = 2 \pi \sqrt{\frac{L}{\vec{g}}}$ is the period, and $\vec{g} > 0$.

We recognize that $\boldsymbol{\left(6\right)}$ is an oscillatory solution, with initial angle ${\vec{\theta}}_{0}$. This holds true for small angles of $\theta$, such as less than ${5}^{\circ}$.

This is the restriction we place onto any simple harmonic oscillator; it works best for small angles of displacement if it's a pendulum, or for small horizontal cartesian displacement if it's a ball and spring.

DISCLAIMER: LONG ANSWER/DERIVATION!

The free-body diagram looks like this: where $\vec{g} > 0$ and horizontally-outwards and down are positive (thus, horizontally-inwards and up are negative).

Since this is the main driving force, we set it equal to $m \vec{a}$ and get:

$- m \vec{g} \sin \vec{\theta} = m \vec{a} = m \frac{{d}^{2} \vec{x}}{{\mathrm{dt}}^{2}}$

where $\frac{{d}^{2} \vec{x}}{{\mathrm{dt}}^{2}}$ is the second derivative of position with respect to time, and is the definition of acceleration.

Cancel out the mass to get:

$\frac{{d}^{2} \vec{x}}{{\mathrm{dt}}^{2}} = - \vec{g} \sin \vec{\theta}$ $\text{ } \boldsymbol{\left(1\right)}$

Now, we need to figure out what the acceleration is. We can start by knowing that the horizontal displacement is:

$\vec{x} = L \sin \vec{\theta}$

where $L$ is the length of the pendulum.

We take $\sin \theta \approx \theta$ for small angles of displacement. Therefore, taking the second derivative of this with respect to time gives:

$\frac{{d}^{2} \vec{x}}{{\mathrm{dt}}^{2}} \approx L \frac{{d}^{2} \vec{\theta}}{{\mathrm{dt}}^{2}}$ $\text{ } \boldsymbol{\left(2\right)}$

Substituting $\boldsymbol{\left(2\right)}$ into $\boldsymbol{\left(1\right)}$, we get:

$L \frac{{d}^{2} \vec{\theta}}{{\mathrm{dt}}^{2}} \approx - \vec{g} \vec{\theta}$

$\frac{{d}^{2} \vec{\theta}}{{\mathrm{dt}}^{2}} = - \frac{\vec{g}}{L} \vec{\theta}$

$\textcolor{g r e e n}{\frac{{d}^{2} \vec{\theta}}{{\mathrm{dt}}^{2}} + \frac{\vec{g}}{L} \vec{\theta} = 0}$ $\text{ } \boldsymbol{\left(3\right)}$

This is the differential equation for a simple oscillator. This is called a linear second-order ordinary differential equation.

• If you do not know differential equations, you can skip all the way to equation $\boldsymbol{\left(6\right)}$, the result.
• If you do know differential equations, then you should read this.

For this, we can assume a solution of the form:

$\vec{\theta} \left(t\right) = {e}^{r t}$

and we find:

${r}^{2} {\cancel{{e}^{r t}}}^{\ne 0} + \frac{\vec{g}}{L} {\cancel{{e}^{r t}}}^{\ne 0} = 0$

${r}^{2} + \frac{\vec{g}}{L} = 0$

$r = \pm i \sqrt{\frac{\vec{g}}{L}}$

We can then use Euler's formula, ${e}^{i t} = \cos t + i \sin t$, to write a linear combination of the assumed solution and rewrite it in terms of sines and cosines in the real domain:

$\vec{\theta} \left(t\right) = {c}_{1} {e}^{i \sqrt{\frac{\vec{g}}{L}} t} + {c}_{2} {e}^{- i \sqrt{\frac{\vec{g}}{L}} t}$

$= {c}_{1} \left[\cos \left(\sqrt{\frac{\vec{g}}{L}} t\right) + i \sin \left(\sqrt{\frac{\vec{g}}{L}} t\right)\right] + {c}_{2} \left[\cos \left(\sqrt{\frac{\vec{g}}{L}} t\right) - i \sin \left(\sqrt{\frac{\vec{g}}{L}} t\right)\right]$

$= \left({c}_{1} + {c}_{2}\right) \cos \left(\sqrt{\frac{\vec{g}}{L}} t\right) + \left(i {c}_{1} - i {c}_{2}\right) \sin \left(\sqrt{\frac{\vec{g}}{L}} t\right)$

For convenience, call ${c}_{1} + {c}_{2} = A$ and $i {c}_{1} - i {c}_{2} = B$ (arbitrary constants) to get:

$\vec{\theta} \left(t\right) = A \cos \left(\sqrt{\frac{\vec{g}}{L}} t\right) + B \sin \left(\sqrt{\frac{\vec{g}}{L}} t\right)$ $\text{ } \boldsymbol{\left(4\right)}$

The pendulum has certain initial conditions:

• $\vec{\theta} \left(t = 0\right) = {\vec{\theta}}_{0}$, i.e. there is a starting angle ${\vec{\theta}}_{0}$ (we set clockwise to be positive).
• $\frac{\mathrm{dv} e c \theta \left(t = 0\right)}{\mathrm{dt}} = 0$, i.e. the pendulum starts at rest and thus does not have a rate of change for the angle yet at time zero.

Apply the second initial condition by taking the first derivative, and then setting the time equal to zero:

$\frac{\mathrm{dv} e c \theta}{\mathrm{dt}} = - A \sqrt{\frac{\vec{g}}{L}} \sin \left(\sqrt{\frac{\vec{g}}{L}} t\right) + B \sqrt{\frac{\vec{g}}{L}} \cos \left(\sqrt{\frac{\vec{g}}{L}} t\right)$

At $t = 0$, we have:

${\cancel{- A \sqrt{\frac{\vec{g}}{L}} \sin \left(\sqrt{\frac{\vec{g}}{L}} \cdot 0\right)}}^{0} + B \sqrt{\frac{\vec{g}}{L}} {\cancel{\cos \left(\sqrt{\frac{\vec{g}}{L}} \cdot 0\right)}}^{1} = 0$

The only way for this condition to hold true is if $B = 0$, so from $\boldsymbol{\left(4\right)}$, we get:

$\vec{\theta} \left(t\right) = A \cos \left(\sqrt{\frac{\vec{g}}{L}} t\right)$ $\text{ } \boldsymbol{\left(5\right)}$

From the first initial condition, we have:

$\vec{\theta} \left(0\right) = {\vec{\theta}}_{0} = A {\cancel{\cos \left(\sqrt{\frac{\vec{g}}{L}} \cdot 0\right)}}^{1}$

Therefore, $A = {\vec{\theta}}_{0}$, and we get the solution:

$\textcolor{b l u e}{\vec{\theta} \left(t\right) = {\vec{\theta}}_{0} \cos \left(\sqrt{\frac{\vec{g}}{L}} t\right)}$ $\text{ } \boldsymbol{\left(6\right)}$

We recognize that $\boldsymbol{\left(6\right)}$ is an oscillatory solution, with initial angle ${\vec{\theta}}_{0}$. This holds true for small angles of $\theta$, such as less than ${5}^{\circ}$.

This is the restriction we place onto any simple harmonic oscillator; it works best for small angles of displacement if it's a pendulum, or for small horizontal cartesian displacement if it's a ball and spring.

Lastly, we need to find the period $T$ of the simple pendulum. Normally we write that the simple harmonic solution is of the form:

$\vec{\theta} \left(t\right) = A \cos \left(\omega t\right)$,

where $\omega$ is the angular frequency in $\text{rad/s}$.

In this case, we must have that:

$\omega = \sqrt{\frac{\vec{g}}{L}}$

We also have the relationship that $\frac{\omega}{2 \pi} = f$, the frequency in ${\text{s}}^{- 1}$ (the $2 \pi$ is the distance of 1 revolution in radians), and that $T = \frac{1}{f}$ is the period in $\text{s}$. Therefore:

$\omega = 2 \pi f = \frac{2 \pi}{T}$

Or, we have that:

$T = \frac{2 \pi}{\omega}$

and therefore:

$\textcolor{b l u e}{T = 2 \pi \sqrt{\frac{L}{\vec{g}}}}$