Question

The function $sin\omega t-cos\omega t$ represents

• A. a simple harmonic motion with a period $\frac{\pi }{\omega }$.
• B. a simple harmonic motion with a period $\frac{2\pi }{\omega }$.
• C. a periodic, but not simple harmonic motion with a period $\frac{\pi }{\omega }$.
• D. a periodic, but not simple harmonic motion with a period $\frac{2}{\pi }\omega$.

Answer 1

The correct option is B a simple harmonic motion with a period $\frac{2\pi }{\omega }$.

$x=\sin \omega t-\cos \omega t$

$⟹x=2\cos \omega t=2\sin (\omega t+\frac{\pi }{2})[\sin C-\sin D=2\cos (\frac{C+D}{2}\left)\sin \right(\frac{C-D}{2}\left)\right]$

On comparing this with equation of $SHM:$ $x=A\sin (\omega t+\varphi )$

$⟹A=2\omega \text{(Angular frequency)}$

$⟹T=\frac{2\pi }{\omega }$