LEARNING OBJECTIVE: 7. 3 Describe the displacement, velocity, and acceleration of an object exhibiting SHM.
As described in previous section the force and acceleration of a body of mass m performing SHM can be related as :
$F = m a = – kx$, hence
$a = – \frac{k}{m} x = – 4 \pi^2f^2 x $
This means we can find the frequency of the vibration from the force equation as follows:
$f = \frac{1}{2 \pi} \sqrt{\frac{k}{m}} $
Given, time period T = 1/f, we can also find time period as:
$T = 2 \pi \sqrt{\frac{m}{k}}$
This is a remarkable result.
The period of oscillation does not depend upon the amplitude. But only on the ratio of mass and force constant alone.
SHM Feature
For SHM we know that, $x = A Cos \omega t$
Differentiating with respect to time we get:
$V = \frac{dx}{dt} = -A \omega Sin \omega t$
Differentiating again we get:
$ \frac{d^2x}{dt^2} = -\omega^2 A Cos \omega t = – \omega^2 x $
Also, the acceleration is:
$\frac{d^2x}{dt^2} = – \frac{k}{m} x$
This means
$\omega = \sqrt{\frac{k}{m}}$
We can now use Newton’s law format of SHM given by:
$F = – m \omega^2x$
to determine $\omega$, directly from equation of motion obtained for a system, where force is proportional to x. Once $\omega$ is known we can then easily calculate f and T using equations derived earlier.
