Question

Which of the following functions of time represent $\left(A\right)$ simple harmonic, $\left(B\right)$ periodic but not simple harmonic, and $\left(C\right)$ non-periodic motion? Give period for each case of periodic motion ( $\omega$ is any positive constant):
(a) $\sin \omega t-\cos \omega t$
(b) ${\sin }^3\omega t$
(c) $3\cos (\pi /4-2\omega t)$
(d) $\cos \omega t+\cos 3\omega t+\cos 5\omega t$
(e) $exp(-{\omega }^2{t}^2)$
(f) $1+\omega t+{\omega }^2{t}^2$

Answer 1

(a) SHM
The given function is:

$sin\omega t-cos\omega t$

$=\sqrt{2}[\frac{1}{\sqrt{2}}sin\omega t-\frac{1}{\sqrt{2}}cos\omega t]$

$=\sqrt{2}[sin\omega t\times cos\frac{\pi }{4}-cos\omega t\times sin\frac{\pi }{4}]$
$=\sqrt{2}sin(\omega t-\frac{\pi }{4})$

This function represents SHM as it can be written in the form: a $sin(\omega t+\varphi )$ Its period is: $2\pi /\omega$.

(b) Periodic but not SHM
The given function is:

$si{n}^3\omega t=1/4[3sin\omega t-sin3\omega t]$
The terms sin ωt and sin 3ωt individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.
Its period is: $2\pi /\omega$ (LCM of time periods).
(c) SHM
The given function is:

$3cos[\frac{\pi }{4}-2\omega t]$

$=-3cos[2\omega t-\frac{\pi }{4}]$

This function represents simple harmonic motion because it can be written in the form: a $cos(\omega t+\varphi )$. Its period is: $2\pi /2\omega =\pi /\omega$
(d) Periodic, but not SHM
The given function is $cos\omega t+cos3\omega t+cos5\omega t$. Each individual cosine function represents SHM. However, the superposition of three simple harmonic motions is periodic, but not simple harmonic.

Its period is LCM of the period of the three sinusoids = $2\pi /\omega$

(e) Non-periodic motion
The given function exp$(-{\omega }^2{t}^2)$ is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.

(f) The given function $1+\omega t+{\omega }^2{t}^2$ is non-periodic.