Two particle executing S.H.M. of same amplitude of 20 ,cm with same period along the same line about same equilibrium position. The maximum distance between

Q: Two particle executing $S.H.M.$ of same amplitude of $20 \,cm$ with same period along the same line about same equilibrium position. The maximum distance between the two is $20 \,cm$. Their phase difference in radian is equal to

a (a) $x_1=A \sin \left(\omega t+\phi_1\right)$ $x_2=A \sin \left(\omega t+\phi_2\right)$ $x_1-x_2=A \sin \left(\omega t+\phi_1\right)-A \sin \left(\omega t+\phi_2\right)$ $20=2 \times 20 \sin \left(\frac{\phi_1-\phi_2}{2}\right) \cdot \cos \left[\omega t+\left(\frac{\phi_1+\phi_2}{2}\right)\right]$ $\frac{1}{2}=\sin \left(\frac{\phi_1-\phi_2}{2}\right) \cdot \cos \left(\omega t+\left(\frac{\phi_1+\phi_2}{2}\right)\right) \text { for maximum value. } \Rightarrow \frac{\phi_1-\phi_2}{2}=\frac{\pi}{6} \Rightarrow \phi_1-\phi_2=\frac{\pi}{3}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

Two particle executing $S.H.M.$ of same amplitude of $20 \,cm$ with same period along the same line about same equilibrium position. The maximum distance between the two is $20 \,cm$. Their phase difference in radian is equal to

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

Medium

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

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