Question

Two simple harmonic motions are given by $y_1=a\sin [\left(\frac{\pi }{2}\right)t+\varphi ]$ and $y_2=b\sin [\left(\frac{2\pi }{3}\right)t+\varphi ]$. The phase difference between these after 1 s is:

• A. zero
• B. $\pi /2$
• C. $\pi /4$
• D. $\pi /6$

Answer 1

The correct option is D $\pi /6$

At t = 1 s
${\varphi }_1=\frac{\pi }{2}+\varphi$
and ${\varphi }_2=\frac{2\pi }{3}+\varphi$
$\therefore$ $\Delta \varphi ={\varphi }_2-{\varphi }_1=\frac{\pi }{6}$