Q.Show that the motion of a particle represented by $y = \sin\omega t - \cos\omega t$ is simple harmonic with a period of $2\pi/\omega$.

The key idea is that any linear combination of $\sin\omega t$ and $\cos\omega t$ with the same angular frequency $\omega$ represents simple harmonic motion (SHM) with that same frequency. The period is determined solely by $\omega$.

First, rewrite the given equation in the standard SHM form $y = A\sin(\omega t + \phi)$ or $y = A\cos(\omega t + \phi)$. Use the identity:

Recognise that $\frac{1}{\sqrt{2}} = \cos\frac{\pi}{4}$ and $\frac{1}{\sqrt{2}} = \sin\frac{\pi}{4}$, so:

The key insight is that simple harmonic motion is defined by a restoring force proportional to displacement, which mathematically gives a sinusoidal function of time with a single angular frequency. Here, both $\sin\omega t$ and $\cos\omega t$ oscillate with the same $\omega$, so their sum (or difference) must also oscillate with that same frequency — but we need to check that it can be written as a single sine or cosine wave.

A common trick: any expression of the form $A\sin\omega t + B\cos\omega t$ can be rewritten as $R\sin(\omega t + \phi)$ or $R\cos(\omega t - \phi)$, where $R = \sqrt{A^2 + B^2}$. This is exactly the form of simple harmonic motion, where the displacement varies sinusoidally with time.

Let’s work through it.

1. Rewrite the given expression in standard SHM form.

   We have $y = \sin\omega t - \cos\omega t$.

   Compare with $R\sin(\omega t + \phi) = R(\sin\omega t \cos\phi + \cos\omega t \sin\phi)$.

   Matching coefficients:

   $\sin\omega t$ term: $R\cos\phi = 1$

   $\cos\omega t$ term: $R\sin\phi = -1$

   (Notice the sign: the $\cos\omega t$ term in the expansion has coefficient $R\sin\phi$, and we need it to equal $-1$.)

2. Find $R$ and $\phi$.

   Square and add: $R^2(\cos^2\phi + \sin^2\phi) = 1^2 + (-1)^2 = 2$, so $R = \sqrt{2}$.

   Then $\tan\phi = \frac{R\sin\phi}{R\cos\phi} = \frac{-1}{1} = -1$, so $\phi = -\pi/4$ (or $3\pi/4$, but we take the principal value that satisfies both signs).

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

3. Confirm it is simple harmonic.

   The displacement $y$ is a pure sine function of $(\omega t - \pi/4)$. This satisfies $d^2y/dt^2 = -\omega^2 y$, which is the hallmark of SHM — acceleration proportional to negative displacement. So the motion is indeed simple harmonic. …

Step 1: Write $y=\sin\omega t-\cos\omega t$ in the form $R\sin(\omega t+\phi)=R(\sin\omega t\cos\phi+\cos\omega t\sin\phi)$ and match coefficients: $R\cos\phi=1$ (from the $\sin\omega t$ term) and $R\sin\phi=-1$ (from the $\cos\omega t$ term).

Step 2: Square and add: $R^2=1^2+(-1)^2=2\Rightarrow R=\sqrt2$; and $\tan\phi=-1\Rightarrow\phi=-\pi/4$.

Step 3: So $y=\sqrt2\sin\!\left(\omega t-\dfrac{\pi}{4}\right)$ — a single sinusoid of angular frequency $\omega$, hence $\dfrac{d^2y}{dt^2}=-\omega^2y$, exactly the SHM condition. …