Question

Two particles are executing S.H.M of the same amplitude of $20\text{cm}$ with the same period along the same line about the same equilibrium position. The maximum distance between the two is $20\text{cm}$. Their phase difference (in radians) is equal to

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{2}$
• C. $\frac{2\pi }{3}$
• D. $\frac{4\pi }{5}$

Answer 1

The correct option is A $\frac{\pi }{3}$

Given that
Amplitude of both the particles, $A=20\text{cm}$
Let the equation of motion of the two particles be
$x_1=A\sin (\omega t+{\varphi }_1)$
$x_2=A\sin (\omega t+{\varphi }_2)$
The distance between the two particles is
$x_1-x_2=A\sin (\omega t+{\varphi }_1)-A\sin (\omega t+{\varphi }_2)$
Maximum distance between particles $=20\text{cm}$ [given]
$\Rightarrow 20=2\times 20\sin \left(\frac{{\varphi }_1-{\varphi }_2}{2}\right).\cos [\omega t+\left(\frac{{\varphi }_1+{\varphi }_2}{2}\right)]$
$\Rightarrow \frac{1}{2}=\sin \left(\frac{{\varphi }_1-{\varphi }_2}{2}\right).\cos (\omega t+\left(\frac{{\varphi }_1+{\varphi }_2}{2}\right))...\left(i\right)$
Equation $\left(i\right)$ will attain maximum value when the term $\cos (\omega t+(\frac{{\varphi }_1+{\varphi }_2}{2}\left)\right)$ becomes $1$
Then, phase difference ${\sin }^{-1}\left(\frac{1}{2}\right)=\left(\frac{{\varphi }_1-{\varphi }_2}{2}\right)$
$\Rightarrow \frac{{\varphi }_1-{\varphi }_2}{2}=\frac{\pi }{6}\Rightarrow {\varphi }_1-{\varphi }_2=\frac{\pi }{3}$