Question

The function of time representing a simple harmonic motion with a period of $\frac{\mathrm{\pi }}{\omega }$ is

• A. $\cos \left(\omega t\right)+\cos \left(2\omega t\right)+\cos \left(3\omega t\right)$
• B. $3\cos \left(\frac{\mathrm{\pi }}{4}-2\omega t\right)$
• C. ${\sin }^2\omega t$
• D. $\sin \left(\omega t\right)+\cos \left(\omega t\right)$

Answer 1

The correct option is B

$3\cos \left(\frac{\mathrm{\pi }}{4}-2\omega t\right)$

Step 1: Given data

Period of simple harmonic motion, $T=\frac{\pi }{\omega }\dots \dots \dots \dots \dots \dots \left(1\right)$

Step 2: Formula used

The general simple harmonic equation:

$y=A\cos \left(\omega \text{'}t\pm \varphi \right)$

Where $y$ is displacement, $A$ is amplitude.

Step 3: Finding the right function

We know that for simple harmonic motion,

$T=\frac{2\mathrm{\pi }}{{\omega }^{\text{'}}}\dots \dots \dots \dots \dots \dots \left(2\right)$

Where $\omega \text{'}=$the angular frequency of SHM and $T=$time period of SHM

Now compare the given simple harmonic equation to the Simple harmonic equation $\left(1\right)$ and equation$\left(2\right)$,

Therefore,

$\begin{array}{rcl}\frac{\pi }{\omega }& =& \frac{2\pi }{{\omega }^{\text{'}}}\\ {\omega }^{\text{'}}& =& 2\omega \end{array}$

Taking function, $3\cos \left(2\omega t-\frac{\mathrm{\pi }}{4}\right)$ which can be written as $3\cos \left(\frac{\mathrm{\pi }}{4}-2\omega t\right)$.

Thus, among all the given functions, only $3\cos \left(\frac{\mathrm{\pi }}{4}-2\omega t\right)$ has the angular frequency of $2\omega$.

Hence, option B is the correct answer.