Question

Two equation of two simple harmonic motion are $x=a\sin (\omega t-\alpha )$ and $y=b\cos (\omega t-\alpha )$. The phase difference between the two is:-

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

Answer 1

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

$x=a\sin \left(\omega t-\alpha \right)=a\cos \left(\omega t-\alpha -\frac{\pi }{2}\right)$

Another equation is

$y=b\cos \left(\omega t-\alpha \right)$

$\therefore$ The phase difference $=(x-y)$ at $t=0$

$\begin{array}{c}a\cos \left(\omega .0-\alpha -\frac{\pi }{2}\right)-b\cos \left(\omega .0-\alpha \right)\\ =\frac{\pi }{2}={90}^0\end{array}$