# Simple Harmonic Motion

• February 17th 2010, 10:30 AM
Yehia
Simple Harmonic Motion
Regarding Simple Harmonic Motion, and a spring moving up and down, when the displacement is largest apparently the acceleration is also maximum. But WHY? When the distance is biggest, isnt speed 0 so the system is also not accelerating?!

So why is acceleration maximum at maximum displacement?!

help much appreciated! thanks!
• February 17th 2010, 12:27 PM
Hello Yehia
Quote:

Originally Posted by Yehia
Regarding Simple Harmonic Motion, and a spring moving up and down, when the displacement is largest apparently the acceleration is also maximum.

You are quite right: when the displacement is largest, the acceleration is also a maximum.
Quote:

But WHY?
Because this is the basic characteristic of SHM - the acceleration is proportional to the displacement. In mathematical terms, this is represented by the fundamental equation of SHM:
$\underbrace{\ddot{x}}_{\text{acceleration }} \underbrace{= -\omega^2}_{\text{ is proportional to }}\underbrace{x}_{\text{ displacement}}$
The minus sign in this equation means that the acceleration is always in the opposite direction to the displacement. In other words, the acceleration is always directed towards the centre of the motion.
Quote:

When the distance is biggest, isnt speed 0 so the system is also not accelerating?!
Not at all! You're confusing speed with acceleration. It is perfectly possible for a body to have zero speed and yet be accelerating. This is exactly what happens at the extremities of the simple harmonic motion.

If you want another illustration of zero speed, but non-zero acceleration, consider a ball that is thrown vertically upwards. As you know, it is accelerating downwards with a constant gravitational acceleration $g$. During the upward phase, gravity is slowing the ball down. At the top of its flight, its velocity is zero; then gravity starts to increase the speed for the downward part of the motion. At the top, the speed is zero, but the acceleration is a constant $g$ downwards for the whole flight!

• February 17th 2010, 12:44 PM
Yehia
Quote:

$\underbrace{\ddot{x}}_{\text{acceleration }} \underbrace{= -\omega^2}_{\text{ is proportional to }}\underbrace{x}_{\text{ displacement}}$
If you want another illustration of zero speed, but non-zero acceleration, consider a ball that is thrown vertically upwards. As you know, it is accelerating downwards with a constant gravitational acceleration $g$. During the upward phase, gravity is slowing the ball down. At the top of its flight, its velocity is zero; then gravity starts to increase the speed for the downward part of the motion. At the top, the speed is zero, but the acceleration is a constant $g$ downwards for the whole flight!