Question

Two particles are in S.H.M, of same amplitude 'a', same frequency 'f', along the same straight line about the same mean position. The maximum distance between the two particles is $a\sqrt{2}$. The initial phase difference between the particles is :

• A. $\frac{\pi }{2}$
• B. zero
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is A $\frac{\pi }{2}$

$x_1=Asin(wt+{\varphi }_1)$

$x_2=Asin(wt+{\varphi }_2)$

$|x_1-x_2|=2Asin[wt+\frac{{\varphi }_1+{\varphi }_2}{2}]cos\frac{{\varphi }_1-{\varphi }_2}{2}$

$|x_1-x_2{|}_{max}\Rightarrow sin[wt+\frac{{\varphi }_1+{\varphi }_2}{2}]$

$\Rightarrow a\sqrt{2}=2a\times 1\times cos\left[\frac{{\varphi }_1-{\varphi }_2}{2}\right]$

$\Rightarrow \frac{{\varphi }_1-{\varphi }_2}{2}=\frac{\pi }{4}\Rightarrow \Delta \varphi =\frac{\pi }{2}$

Hence "A"