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Two particles 'A' and 'B' perform S.H.M., starting from mean position have periodic the ' $T$ ' and $3 T / 2$ respectively. The phase difference between particles A and B when particle A completes one oscillation is
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The correct answer is:
$\frac{2 \pi}{3}$
Equation of motion of the practical are
$\mathrm{X}_1=\mathrm{A}_1 \sin \frac{2 \pi}{\mathrm{T}_1} \mathrm{t}$ and $\mathrm{X}_2=\mathrm{A}_2 \sin \frac{2 \pi}{\mathrm{T}_2} \mathrm{t}$
$\therefore$ Phase difference $\Delta \phi=\left(\frac{2 \pi}{\mathrm{T}_1}-\frac{2 \pi}{\mathrm{T}_2}\right) \mathrm{t}$
$\begin{aligned} & =\left(\frac{2 \pi}{T}-\frac{2 \pi}{3 T / 2}\right) t \\ & \text { At } t=T \\ & \Delta \phi=\left(2 \pi-\frac{2 \times 2 \pi}{3}\right) \frac{T}{T}=\frac{2 \pi}{3}\end{aligned}$
$\mathrm{X}_1=\mathrm{A}_1 \sin \frac{2 \pi}{\mathrm{T}_1} \mathrm{t}$ and $\mathrm{X}_2=\mathrm{A}_2 \sin \frac{2 \pi}{\mathrm{T}_2} \mathrm{t}$
$\therefore$ Phase difference $\Delta \phi=\left(\frac{2 \pi}{\mathrm{T}_1}-\frac{2 \pi}{\mathrm{T}_2}\right) \mathrm{t}$
$\begin{aligned} & =\left(\frac{2 \pi}{T}-\frac{2 \pi}{3 T / 2}\right) t \\ & \text { At } t=T \\ & \Delta \phi=\left(2 \pi-\frac{2 \times 2 \pi}{3}\right) \frac{T}{T}=\frac{2 \pi}{3}\end{aligned}$
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