Match Periodic Functions with Time Period and SHM Behavior

To determine the correct match between List - I and List - II, we analyze each function given in List - I:

1. Analysis of Function A: $y = \sin^2 \omega t$

Using the trigonometric power-reduction identity, we can rewrite $y$ as: $y = \frac{1 - \cos(2\omega t)}{2} = \frac{1}{2} - \frac{1}{2}\cos(2\omega t)$

Differentiating with respect to time $t$: $\frac{dy}{dt} = \omega \sin(2\omega t)$ $\frac{d^2 y}{dt^2} = 2\omega^2 \cos(2\omega t)$

Since $\cos(2\omega t) = 1 - 2y$, substituting this back gives: $\frac{d^2 y}{dt^2} = 2\omega^2 (1 - 2y) = -4\omega^2 \left(y - \frac{1}{2}\right)$

This differential equation is of the standard SHM form $\frac{d^2 y}{dt^2} = -\omega'^2 (y - y_0)$ where $\omega' = 2\omega$ and the mean position is $y_0 = \frac{1}{2}$.

Thus, the motion is Simple Harmonic Motion (SHM) with a time period: $T = \frac{2\pi}{\omega'} = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$

Hence, A matches with III.

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2. Analysis of Function B: $y = \sin^3(2\omega t)$

Using the identity $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$, we get: $\sin^3\theta = \frac{3\sin\theta - \sin(3\theta)}{4}$

For $\theta = 2\omega t$: $y = \frac{3}{4}\sin(2\omega t) - \frac{1}{4}\sin(6\omega t)$

This is a linear combination of two harmonic functions with different angular frequencies ($\omega_1 = 2\omega$ and $\omega_2 = 6\omega$). Because it contains multiple frequencies, it cannot be represented as a single sine or cosine term, so it is not SHM.

The fundamental angular frequency is $\omega_0 = \text{gcd}(2\omega, 6\omega) = 2\omega$. The time period of the periodic motion is: $T = \frac{2\pi}{\omega_0} = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$

Hence, B matches with I.

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3. Analysis of Function C: $y = \sin(\omega t) + \cos(\pi \omega t)$

The angular frequencies of the two constituent waves are: $\omega_1 = \omega \quad \text{and} \quad \omega_2 = \pi \omega$

Taking their ratio: $\frac{\omega_1}{\omega_2} = \frac{\omega}{\pi \omega} = \frac{1}{\pi}$

Since $\frac{1}{\pi}$ is an irrational number, there exists no common time period $T$ such that $y(t+T) = y(t)$ for all $t$. Therefore, the function is non-periodic.

Hence, C matches with IV.

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4. Analysis of Function D: $y = \cos \omega t + \cos 2\omega t$

This function is a sum of two harmonic oscillations with angular frequencies $\omega_1 = \omega$ and $\omega_2 = 2\omega$. Since it contains more than one frequency, it is not SHM.

The ratio of frequencies $\frac{\omega_1}{\omega_2} = \frac{1}{2}$ is a rational number, so the motion is periodic. The fundamental angular frequency is $\omega_0 = \text{gcd}(\omega, 2\omega) = \omega$.

The time period is: $T = \frac{2\pi}{\omega}$

Hence, D matches with II.

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Conclusion:

• A $\rightarrow$ III
• B $\rightarrow$ I
• C $\rightarrow$ IV
• D $\rightarrow$ II

This corresponds to Option A.