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Question: Answered & Verified by Expert
Show that the motion of a particle represented by $y=\sin \omega t-\cos \omega t$ is simple harmonic with a period of $2 \pi / \omega$.
PhysicsOscillations
Solution:
1532 Upvotes Verified Answer
When we convert the given combination of two simple harmonic functions to a single harmonic (sine or cosine) function it can be written uniquely in the form of $\cos \left(\frac{2 \pi}{T} t+\phi\right)$ or $\sin \left(\frac{2 \pi}{T} t+\phi\right)$
As given that the displacement function
$$
\begin{aligned}
y &=\sin \omega t-\cos \omega t \\
&=\sqrt{2}\left(\frac{1}{\sqrt{2}} \cdot \sin \omega t-\frac{1}{\sqrt{2}} \cdot \cos \omega t\right)
\end{aligned}
$$
$$
\begin{aligned}
&=\sqrt{2}\left[\cos \left(\frac{\pi}{4}\right) \cdot \sin \omega t-\sin \left(\frac{\pi}{4}\right) \cdot \cos \omega t\right] \\
&=\sqrt{2}\left[\sin \left(\omega t-\frac{\pi}{4}\right)\right]=\sqrt{2}\left[\sin \left(\omega t-\frac{\pi}{4}\right)\right]
\end{aligned}
$$
Comparing with standard SHM equation
$$
y=a \sin \left(\frac{2 \pi}{T} t+\phi\right)
$$
then we get
$$
\omega=\frac{2 \pi}{T} \text { or } T=\frac{2 \pi}{\omega}
$$
So, clearly, the function represents SHM with a period $T=\frac{2 \pi}{\omega}$. Hence proved.

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