MCQ
Two bodies performing $SHM$ have same amplitude and frequency. Their phases at a certain instant are as shown in the figure. The phase difference between them is
- A$\pi $
- B$\frac{{2\pi }}{3}$
- ✓$\frac{{\pi }}{2}$
- D$\frac{{3\pi }}{4}$
✓
Answer
Correct option: C.
$\frac{{\pi }}{2}$
c
Displacement from mean position
Displacement from mean position
$=\mathrm{A}-\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \mathrm{A}=\frac{\mathrm{A}}{\sqrt{2}}$
So phase of the particle which is going towards
$+x$ is $\phi_{1}=\frac{\pi}{4}$
and for other particle $\quad \phi_{2}=2 \pi-\pi / 4$
$=\frac{7 \pi}{4}$
Phase difference $=\frac{7 \pi}{4}-\frac{\pi}{4}=\frac{6 \pi}{4}=\frac{3 \pi}{2}$ or $\frac{\pi}{2}$
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