Two harmonic oscillators A and B moving in the x direction according to XA=X0sintt0 and XB=X0sin(tt0+π2) respectively, starting from t=0 . Take x0=1m. t0=1s , then magnitudes of their relative velocities in m/s at time t=π3s is

XA=sint⇒V¯A=costi¯XB=cost,V¯B=−sinti¯|V¯A−V¯B|=3+12 (∵t=π3)

Two harmonic oscillators A and B moving in the x direction according to XA=X0sintt0 and XB=X0sin(tt0+π2) respectively, starting from t=0 . Take x0=1m. t0=1s , then magnitudes of their relative velocities in m/s at time t=π3s is

XA=sint⇒V¯A=costi¯XB=cost,V¯B=−sinti¯|V¯A−V¯B|=3+12 (∵t=π3)

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Two harmonic oscillators A and B moving in the x direction according to $X_{A} = X_{0} \sin \frac{t}{t_{0}}$ and $X_{B} = X_{0} \sin (\frac{t}{t_{0}} + \frac{π}{2})$ respectively, starting from $t = 0$ . Take $x_{0} = 1 m$ . $t_{0} = 1 s$ , then magnitudes of their relative velocities $i n m/s$ at time $t = \frac{π}{3} s$ is

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$\frac{\sqrt{3} + 1}{2}$

$\frac{\sqrt{3} - 1}{2}$

$\frac{\sqrt{3} - 1}{4}$

$\frac{- \sqrt{3} + 1}{2}$

answer is A.

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$X_{A} = \sin t \Rightarrow {\bar{V}}_{A} = \cos t \bar{i}$

$X_{B} = \cos t, {\bar{V}}_{B} = - \sin t \bar{i}$

$| {\bar{V}}_{A} - {\bar{V}}_{B} | = \frac{\sqrt{3} + 1}{2}$ $(∵ t = \frac{π}{3})$

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Two harmonic oscillators A and B moving in the x direction according to XA=X0sintt0 and XB=X0sin(tt0+π2) respectively, starting from t=0 . Take x0=1m. t0=1s , then magnitudes of their relative velocities in m/s at time t=π3s is

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