# Amplitude of a pendulum

I have a pendulum drawn back at a height of 20mm from the surface, its length is 800mm and is a 25g mass. I'm asked to calculate the initial amplitude of the pendulum as well as its maximum speed and I do so by firstly calculating the period by $T=2π\sqrt{\frac{L}{g}}$ giving me $1.79s$, I then proceed to use $mgh=0.5mv²$ as energy is conserved in a pendulum ignoring all friction with air and then sub my value of velocity into $v=2πfA$ and get my correct answer. However, looking at another equation for maximum acceleration which is $a=(2πf)²A$ I thought that seeing the object only accelerates under $9.81m/s ²$, I assumed this equation would suffice, however I get the wrong answer. Is there infact a greater maximum acceleration possible meaning that this equation wont work with what I need (The initial amplitude)?

• A pendulum fixed at a point will undergo rotational motion. Is the acceleration the bob experiences really only that of $g = -9.81\frac{m}{s^2}$? – bleuofblue Oct 31 '16 at 20:03
• As far as I know yes. – Jan Oct 31 '16 at 21:09
• Think a little bit more about how the bob will accelerate. If gravity was the only force accelerating the bob, what is causing it to move horizontally at all? What you need to discern is the net acceleration from both gravity, and the string pulling it back. – bleuofblue Oct 31 '16 at 21:17

The amplitude of a pendulum is not a well defined term. It can be measured by horizontal displacement or angular displacement.

When the angular displacement of the bob is $\theta$ radians, the tangential acceleration is $a=-g\sin \theta$ which is "towards" the equilibrium point - ie acceleration is always opposite the direction in which $\theta$ is increasing. (Think of the bob sliding down an inclined plane at angle $\theta$.) The acceleration is greatest when $\theta$ equals the amplitude, and zero when $\theta=0$.

The above formula for $a$ is exact. You have to be careful when using other formulas which use the small angle approximation (SAA) : $\sin\theta \approx \theta$.

Your formula $a \approx -(2\pi f)^2 A$ (note minus sign) is also correct, assuming that $A$ is angular displacement $\theta$, which (using the SAA) varies sinusoidally :
$\theta \approx \theta_0 \sin(2\pi f t)$.
Here $\theta_0$ is the angular amplitude. The linear acceleration is
$a=L\frac{d^2 \theta}{dt^2} \approx -(2\pi f)^2 \theta$.
Note that
$(2\pi f)^2=(2\pi \frac{1}{T})^2 \approx \frac{g}{L}$.
Therefore
$a \approx -g\theta$.

This differs from the equation in the 1st paragraph because it includes the SAA : $\sin\theta \approx \theta$.