Animated Solution for Physics - Oscillations: The function of time representing a simple harmonic motion with a period of $\frac{\pi }{\omega }$ is

Imagine a mass on a spring, bobbing back and forth in a perfect, uninterrupted rhythm. This is the heart of Simple Harmonic Motion (SHM). Mathematically, this elegant dance is captured by a single sine or cosine function: $x(t)=A\sin (\Omega t+\varphi )$ or $x(t)=A\cos (\Omega t+\varphi )$.

The time it takes to complete one full cycle is the time period, $T$, which is intimately tied to the angular frequency, $\Omega$, by the relation $T=\frac{2\pi }{\Omega }$.

In our problem, we are on a quest to find a function that represents SHM with a specific period: $T=\frac{\pi }{\omega }$.

Let's set up our target. If $T=\frac{\pi }{\omega }$, then:

Our mission is clear: we need to find the option that boils down to a single sine or cosine function with an angular frequency of exactly $2\omega$.

Let's put each option under the mathematical microscope.

Option (a): $f(t)=\sin (\omega t)+\cos (\omega t)$

At first glance, this looks like two motions. But using a clever trigonometric trick—multiplying and dividing by $\sqrt{2}$—we can merge them:

This is indeed pure SHM! However, the angular frequency here is just $\omega$, which means its period is $\frac{2\pi }{\omega }$. Close, but not what we are looking for.

Option (b): $f(t)=\cos (\omega t)+\cos (2\omega t)+\cos (3\omega t)$

This is a classic trap. While this function is periodic (it will repeat itself), it is a superposition of three different frequencies. It's like three different springs trying to pull the mass at different rates. This results in a complex wave, not the pure, single-frequency oscillation required for Simple Harmonic Motion.

Option (c): $f(t)={\sin }^2(\omega t)$

This one is sneaky. It doesn't look like standard SHM because of the square. But let's use the half-angle formula:

Surprise! This actually is Simple Harmonic Motion. The angular frequency is $2\omega$, which gives the correct period of $\frac{\pi }{\omega }$. The only catch is the $\frac{1}{2}$ at the front, which means the oscillation happens around a shifted mean position ($x=0.5$) rather than the origin.

Option (d): $f(t)=3\cos \left(\frac{\pi }{4}-2\omega t\right)$

Let's clean this up. Since the cosine function is even, meaning $\cos (-\theta )=\cos (\theta )$, we can flip the terms inside:

This is a beautiful, standard equation for SHM. The amplitude is $3$, the phase constant is $-\frac{\pi }{4}$, and most importantly, the angular frequency is exactly $2\omega$. This gives us the perfect period of $\frac{\pi }{\omega }$.

Both options (c) and (d) yield the correct period. However, option (d) represents the most standard, unshifted form of Simple Harmonic Motion about the origin. In the context of such problems, the pure sinusoidal form without a constant offset is the intended answer. Therefore, option (d) takes the crown!