Question

Two simple harmonic motions are represented by equations $y_1=4\sin (10t+\varphi )$ and $y_2=5\cos \left(10t\right)$. What is the phase difference between their velocities?

• A. $\varphi$
• B. $-\varphi$
• C. $(\varphi +\frac{\pi }{2})$
• D. $(\varphi -\frac{\pi }{2})$

Answer 1

The correct option is D $(\varphi -\frac{\pi }{2})$

Given: $y_1=4\sin (10t+\varphi )$
$y_2=5\cos \left(10t\right)$
We know that velocity$;v=\frac{dy}{dt}$
$\therefore v_1=\frac{d{y}_1}{dt}=40\cos (10t+\varphi )....\left(i\right)$
$v_2=\frac{d{y}_2}{dt}=-50\sin \left(10t\right)v_2=50\cos (10t+\frac{\pi }{2})....\left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$
Phase difference between $v_1$ and $v_2$$=|{\varphi }_2-{\varphi }_1|$
$=(\frac{\pi }{2}-\varphi )$ or $(\varphi -\frac{\pi }{2})$