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Step-by-Step Explanation
Step 1: Examine the function y = sin($\omega t$) – cos($\omega t$)
Combine the sine and cosine terms using a trigonometric identity. We write:
$ y = \sin(\omega t) - \cos(\omega t) $
Notice that
$ \sin(\omega t) - \cos(\omega t)
= \sqrt{2}\left[\dfrac{1}{\sqrt{2}}\sin(\omega t) - \dfrac{1}{\sqrt{2}}\cos(\omega t)\right]
= \sqrt{2}\sin\left(\omega t - \dfrac{\pi}{4}\right). $
This is of the form $ A \sin(\omega t + \phi) $, which describes a simple harmonic motion (SHM) with angular frequency $\omega$. The corresponding time period is
$ T = \dfrac{2\pi}{\omega}. $
Step 2: Examine the function y = sin3($\omega t$)
We can use a trigonometric identity to write:
$ \sin^3(\omega t) = \dfrac{3\sin(\omega t) - \sin(3\omega t)}{4}. $
Although this expression is periodic, it is not of the strict form $ A \sin(\omega t + \phi) $ or $ A \cos(\omega t + \phi) $ and thus does not represent a simple harmonic motion. It represents a periodic motion but not SHM.
Step 3: Examine the function y = 5 cos($\frac{3\pi}{4} - 3\omega t$)
Use the property
$ \cos(-\theta) = \cos(\theta). $
Hence,
$ 5 \cos\left(\dfrac{3\pi}{4} - 3\omega t\right) = 5 \cos\left(3\omega t - \dfrac{3\pi}{4}\right). $
This is of the form $ A \cos(\alpha \, t + \phi) $, which describes SHM with angular frequency $3 \omega$. Therefore, the time period is
$ T = \dfrac{2\pi}{3\omega}. $
Step 4: Examine the function y = 1 + $\omega t$ + $\omega^2 t^2$
This is a polynomial in $t$ and does not repeat its values after equal intervals of time, meaning it is not periodic and thus cannot be a simple harmonic motion. As $t \to \infty$, $y \to \infty$, violating the bounded nature of SHM.
Conclusion
From the analysis, the functions that represent SHM are:
1) $ y = \sin(\omega t) - \cos(\omega t) $
3) $ y = 5 \cos\left(\dfrac{3\pi}{4} - 3\omega t\right). $
Hence, the correct answer is “Only (1) and (3)”.
