MHT CET · Physics · Oscillations
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
- A \(\frac{\pi}{2}\)
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{2 \pi}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 \pi}{3}\)
Step-by-step Solution
Detailed explanation
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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