Question

Two particles are in $SHM$ with same angular frequency and amplitudes $A$ and $2 A $ respectively along same straight line with same mean position. They cross each other at position $\frac{ A }{2}$ distance from mean position in opposite direction. The phase difference between them is:

• A. $\frac{5 \pi}{6}-\sin ^{-1}\left(\frac{1}{4}\right)$
• B. $\frac{\pi}{6}-\sin ^{-1}\left(\frac{1}{4}\right)$
• C. $\frac{5 \pi}{6}-\cos ^{-1}\left(\frac{1}{4}\right)$
• D. $\frac{\pi}{6}-\cos ^{-1}\left(\frac{1}{4}\right)$

Answer 1

Answer: (A)

$\sin \phi_{1}=\frac{ A / 2}{ A }=\frac{1}{2}$
$\phi_{1}=\frac{\pi}{6}$
$\sin \left(\pi-\phi_{2}\right)=\frac{ A / 2}{2 A }=\frac{1}{4}$
$\phi_{2}=\pi-\sin ^{-1}\left(\frac{1}{4}\right)$
Phase difference
$\phi_{2}-\phi_{1}=\frac{5 \pi}{6}-\sin ^{-1}\left(\frac{1}{4}\right)$