Two particles $P$ and $Q$ describe SHM of same amplitude $a$ and same frequency $f$ along the same straight…
Two particles $P$ and $Q$ describe SHM of same amplitude $a$ and same frequency $f$ along the same straight line. The maximum distance between the two particles is $a \sqrt{2}$. The phase difference between the particle is
zero
$\frac{\pi}{2}$
$\frac{\pi}{6}$
$\frac{\pi}{3}$
Solution
Given that, two particles $P$ and $Q$ describes
SHM of same amplitude $a$ and frequency $f$.
Let equation of motion of two particles are
$x_1=a \sin \omega t$ and $x_2=a \sin (\omega t+\phi)$
where, $\phi$ be the phase difference.
$\therefore$ Path difference,
$
\begin{aligned}
& x_2-x_1=a \sin (\omega t+\phi)-a \sin \omega t \\
& x_2-x_1=2 a \sin \left(\frac{\omega t+\phi-\omega t}{2}\right) \cos \left(\frac{\omega t+\phi+\omega t}{2}\right)
\end{aligned}
$
$
\begin{aligned}
& {\left[\text { Using identity } \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cdot \cos \left(\frac{A+B}{2}\right)\right]} \\
& \therefore \quad x_2-x_1=2 a \sin \left(\frac{\phi}{2}\right) \cos \left(\omega t+\frac{\phi}{2}\right)
\end{aligned}
$
The distance $x_2-x_1$ will be maximum when
$
\cos \left(\omega t+\frac{\phi}{2}\right)=1
$
$\therefore \quad$ Maximum distance,
$
\begin{aligned}
x_2-x_1 & =\sqrt{2} a \\
& =2 a \sin \frac{\phi}{2} \\
\Rightarrow \quad \sin \frac{\phi}{2} & =\frac{1}{\sqrt{2}} \Rightarrow \sin \frac{\phi}{2}=\sin 45^{\circ} \\
\Rightarrow \quad \frac{\phi}{2} & =45^{\circ}
\end{aligned}
$
Phase difference, $\phi=90^{\circ}=\frac{\pi}{2}$