Question

The equation of the displacement of two particles making SHM are represented by $y_1=a\,sin(\omega t+\phi)y_2=a\,cos(\omega t)$ The phase difference of the velocities of the two particles is

• A. $\frac{\pi}{2}+\phi$
• B. $-\phi$
• C. $\phi$
• D. $\phi-\frac{\pi}{2}$

Answer 1

Answer: (D)

$y _{1}=a \sin(\omega t +\phi)$
$y _{2}=a \cos(\omega t )$
$y _{1}=a \omega \cos (\omega t +\phi)$
$=a \omega \sin (\omega t +\phi+\pi / 2)$
$y _{2}=- a \omega \sin (\omega t )= a \omega \sin (\pi+\omega t )$
therefore phase difference $=\omega t +\phi+\pi / 2-(\pi+\omega t )$
$=\phi-\pi / 2$