Question

The phase shift of the resultant $y=3\sin \omega t+4\cos \omega t$ is:

• A. ${37}^{\circ }$
• B. ${53}^{\circ }$
• C. ${90}^{\circ }$
• D. None of these

Answer 1

Answer: (B)

The equations can be split up as $y_1=3sin\left(\omega t\right)$ and $y_2=4sin(\frac{\pi }{2}-\omega t)$.
The phase shift is given by, $\delta ={\tan }^{-1}\left(\frac{a_{1}sin{\alpha }_1+a_{2}sin{\alpha }_2}{a_{1}cos{\alpha }_1+a_{2}cos{\alpha }_2}\right)$. Substituting the given values, we get answer as ${53.13}^0$ . Hence, option B is correct.