Q.Which of the following functions of time represent

Key Idea: Frequency And Period

A motion is simple harmonic if it can be written as $A \sin(\omega t + \phi)$ (or $A\cos$) about a fixed equilibrium -- a single sine/cosine with constant angular frequency $\omega$. A function can still represent harmonic motion about a shifted equilibrium if it reduces to a constant plus a single sinusoid.

(1) $\sin\omega t - \cos\omega t$

Rewrite using the identity:

$\sin\omega t - \cos\omega t = \sqrt{2} \sin\left(\omega t - \frac{\pi}{4}\right)$

This is a pure sine wave with amplitude $\sqrt{2}$ and angular frequency $\omega$.

• Simple harmonic motion -- yes.
• Period: $T = \frac{2\pi}{\omega}$.

(2) $\sin^{2}\omega t$

Use $\sin^{2}\theta = \frac{1 - \cos 2\theta}{2}$:

$\sin^{2}\omega t = \frac{1}{2} - \frac{1}{2}\cos 2\omega t$ …

Concept and Intuition

Simple harmonic motion (SHM) is defined by a restoring force proportional to displacement about some equilibrium point, leading to a sinusoidal time dependence of the form $x(t) = A\sin(\omega t + \phi)$ or $x(t) = A\cos(\omega t + \phi)$, possibly shifted by an added constant so the equilibrium is at $x_0 \ne 0$. The motion is periodic with period $T = 2\pi/\omega$, and the function contains only one sinusoidal term at a single frequency (plus, optionally, a constant offset that only shifts the equilibrium, not the character of the motion).

A function can be periodic without being harmonic at all if it repeats after a fixed time interval but cannot be written as a constant plus a single sine/cosine term -- for example, if it contains multiple different frequencies or non-sinusoidal shapes.

The trick is to simplify each given expression using trigonometric identities and see if it collapses into a constant plus a single sine/cosine term.

Step-by-step solution

1. Function (1): $\sin\omega t - \cos\omega t$

We recognise that a linear combination of $\sin$ and $\cos$ with the same frequency can be combined into a single sinusoid.

Recall the identity:

We want to match $\sin\omega t - \cos\omega t$. Comparing coefficients:

• Coefficient of $\sin\omega t$: $R\cos\phi = 1$
• Coefficient of $\cos\omega t$: $-R\sin\phi = -1 \implies R\sin\phi = 1$

Square and add:

Divide the two equations:

Thus:

This is a pure sine wave with amplitude $\sqrt{2}$, angular frequency $\omega$, and phase shift $\pi/4$, oscillating about $x=0$. It represents simple harmonic motion.

The period of SHM is $T = \frac{2\pi}{\omega}$.

2. Function (2): $\sin^{2}\omega t$

Use the power-reduction identity:

Here $\theta = \omega t$, so:

$\sin^{2}\omega t = \frac{1}{2} - \frac{1}{2}\cos 2\omega t$ …

Step 1: A function represents harmonic motion if it reduces to a constant plus a single sinusoid, $C + A\sin(\omega t+\phi)$ (the constant $C$, if any, only shifts the equilibrium point -- it doesn't change whether the motion is harmonic); if it still repeats but can't be written that way, it is periodic but not harmonic.

Step 2: (1) Using $a\sin\theta+b\cos\theta=\sqrt{a^2+b^2}\sin(\theta+\phi)$ with $a=1,b=-1$: $\sin\omega t-\cos\omega t=\sqrt2\sin(\omega t-\pi/4)$ -- a single sinusoid of frequency $\omega$ about $x=0$, so it is SHM with $T=2\pi/\omega$.

Step 3: (2) Apply the double-angle identity $\sin^2\theta=\dfrac{1-\cos2\theta}{2}$: $\sin^2\omega t=\dfrac12-\dfrac12\cos2\omega t$. …