Question

Which one of the following functions of time represents (a) periodic and (b) non-periodic motion? Give the period for each case of periodic motion. [\[\omega \] is any positive constant].
A. \[\dfrac{{2\pi }}{\omega }\]
B. \[\sin \omega t + \cos 2\omega t + \sin 4\omega t\]
C. \[{e^{ - \omega t}}\]
D. \[\log \left( {\omega t} \right)\]

Answer 1

Hint: First of all, we will go for the functions with the trigonometric ratios, as we know that the values of trigonometric ratios repeat in cycles for different values of angles.

Complete answer:
To begin with, we will expand each and every function if there is any period really in there:
A. \[\dfrac{{2\pi }}{\omega }\]
In the above fraction, we can see that this is not a function of time. We can be certain in that ground as there is no time variable \[t\] . To be variable with time, there must contain a time variable. But however, that does not mean that it will be a periodic one. To periodic function, the function should have the same value after a definite interval of time. The fraction \[\dfrac{{2\pi }}{\omega }\] , does not look like a function as there is no variable of time. But however, this looks like a time period for a function such as \[\sin \omega t\] or \[\cos \omega t\] .

B. \[\sin \omega t + \cos 2\omega t + \sin 4\omega t\]
The above function definitely represents a periodic function with the different values of angular frequency.
The term \[\sin \omega t\] has a time period of:
\[T = \dfrac{{2\pi }}{\omega }\]
The term \[\cos 2\omega t\] has time period of:
$T = \dfrac{{2\pi }}{{2\omega }} \\
\Rightarrow T = \dfrac{\pi }{\omega } \\$
The term \[\sin 4\omega t\] has a time period of:
$T = \dfrac{{2\pi }}{{4\omega }} \\
\therefore T = \dfrac{\pi }{{2\omega }} \\$
The last two terms in the function repeats itself after the integral multiple of their period. So, by far each term in the function repeats itself after a time interval of \[T\] . So, the given function is periodic in nature with a period \[T = \dfrac{{2\pi }}{\omega }\] .

C. \[{e^{ - \omega t}}\]
In the above function we can see that the value of the function \[{e^{ - \omega t}}\] decreases with the increase in time. The value of the function won’t het repeated again. Hence, this function is not a period function.

D. \[\log \left( {\omega t} \right)\]
In the above function, we can see that the values of the function \[\log \left( {\omega t} \right)\] increases with the increase in time. The value of the function won’t het repeated again. Hence, this function is not a period function.

Hence, option B is the correct answer.

Note: It is important to remember that a function will only be periodic if we get the same value after a definite interval of time. In the third function, we should know that the value of the function tends to zero when time becomes infinity. Again, in the fourth function, the value of the function is infinity when the time becomes infinity.