Q101 / 125 · BITSAT 2014 · Physics
Two particles $\mathrm{P}$ and $\mathrm{Q}$ describe S.H.M. of same amplitude a, same frequency $\mathrm{f}$ along the same straight line. The maximum distance between the two particles is $\mathrm{a} \sqrt{2}$. The initial phase difference between the particle is $-$
- Azero
- Correct answerB$\pi / 2$
- C$\pi / 6$
- D$\pi \beta$
Solution
$\mathrm{x}_{1}=\mathrm{a} \sin \left(\omega \mathrm{t}+\phi_{1}\right), \mathrm{x}_{2}=\mathrm{a} \sin \left(\omega \mathrm{t}+\phi_{2}\right)$
$\Rightarrow \quad\left|\mathrm{x}_{1}-\mathrm{x}_{2}\right|=2 \mathrm{a} \sin \left(\omega \mathrm{t}+\frac{\phi_{1}+\phi_{2}}{2}\right) \cos \left(\frac{\phi_{1}-\phi_{2}}{2}\right)$
Tomaximize $\left|\mathrm{x}_{1}-\mathrm{x}_{2}\right|: \sin \left(\omega \mathrm{t}+\frac{\phi_{1}+\phi_{2}}{2}\right)=1$
$\Rightarrow \mathrm{a} \sqrt{2}=2 \mathrm{a} \times 1 \times \cos \left(\frac{\phi_{1}-\phi_{2}}{2}\right)$
$\Rightarrow \frac{1}{\sqrt{2}}=\cos \left(\frac{\phi_{1}-\phi_{2}}{2}\right) \Rightarrow \frac{\pi}{4}=\frac{\phi_{1}-\phi_{2}}{2}$
$\Rightarrow \phi_{1}-\phi_{2}=\frac{\pi}{2}$
Hence, the correct answer is option: 'B'.