Question

Two particles are executing SHM. Their displacements at any instant of time t are $x=a\sin \left(\omega t-\varphi \right)$ and $y=b\cos \left(\omega t-\varphi \right)$. The phase difference between them will be:

• A. $0^{\circ }$
• B. $\varphi$
• C. ${90}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is C ${90}^{\circ }$

In the second term, it can be written as $y=bsin(\frac{\pi }{2}-(\omega t-\varphi \left)\right)$.
Comparing it with $x=asin(\omega t-\varphi )$, we see that phase difference is 90 deg. Hence, option C is correct.