The function \(\sin ^{2}(\omega t)\) represents (A) A SHM with periodic time \(\pi / \omega\) (B) A SHM with a periodic time \(2 \pi / \omega\) (C) A periodic motion with periodic time \(\pi / \omega\) (D) A periodic motion with period \(2 \pi / \omega\)

Short Answer

Expert verified
The given function \(\sin^2(\omega t)\) represents a periodic motion (not a Simple Harmonic Motion) with a period of \(2 \pi / \omega\). Therefore, the correct answer is (D) A periodic motion with period \(2 \pi / \omega\).

Step by step solution

01

Recognize the waveform

The given function is \(\sin^2(\omega t)\), which is a trigonometric function squared. It represents a non-negative, periodic motion. However, it is not a Simple Harmonic Motion (SHM) as its shape is not sinusoidal. So options (A) and (B) can be discarded.
02

Evaluate the period of the given function

To find the period of the motion, we need to find the value of \(t\) for which the function repeats its cycle. Let's look for the period of the motion given by \(\sin^2(\omega t)\): We know that the sine function has a period of \(2\pi\), which means: \[\sin(\omega t) = \sin(\omega t + 2 \pi n)\] where \(n\) is an integer. Simply square both sides to find the period of the given function: \[\sin^2(\omega t) = \sin^2(\omega t + 2 \pi n)\] Thus, \(\sin^2(\omega t)\) has the same period as the sine function, which is \(2 \pi\). To find the period concerning \(t\), we need to divide by the coefficient of \(t\) in the argument, which is \(\omega\): Period of \(\sin^2(\omega t) = \dfrac{2 \pi}{\omega}\)
03

Conclusion

Now we know that the given function \(\sin^2(\omega t)\) describes a periodic motion (not an SHM) with period \(2 \pi / \omega\). So the correct answer is: (D) A periodic motion with period \(2 \pi / \omega\).

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