364210 The displacement-time equation of particle executing SHM is: \(x=A \sin (\omega t+\phi)\). At time \(t = 0\), position of the particle is \(x = A/2\) and it is moving along negative \(x\)-direction. Then the angle \(\phi\) can be:

364211 Two particles execute SHM with same amplitudes and same angular frequency on same straight line with same mean position. Given that during oscillation they cross each other in opposite direction when at a distance \(A / 2\) from mean position. Find phase difference in the two \(SHMs\).

364212 Two pendulums have time periods T and \(\frac{{5T}}{4}\).
They start S.H.M. at the same time from the mean position. What will be the phase difference between them after the bigger pendulum
complete one oscillation?

364213 The equation of S.H.M is \(y=a \sin (2 \pi n t+\alpha)\), then its phase at time \(t\) is

364214 Two simple harmonic motions exhibited by two particles \(A\) and \(B\) are given respectively by the following equations.
\({y_1} = a\sin \left[ {\omega t + \frac{\pi }{6}} \right]\)
\({y_2} = a\sin \left[ {\omega t + \frac{{3\pi }}{6}} \right]\)
Phase difference between them is

364210 The displacement-time equation of particle executing SHM is: \(x=A \sin (\omega t+\phi)\). At time \(t = 0\), position of the particle is \(x = A/2\) and it is moving along negative \(x\)-direction. Then the angle \(\phi\) can be:

364211 Two particles execute SHM with same amplitudes and same angular frequency on same straight line with same mean position. Given that during oscillation they cross each other in opposite direction when at a distance \(A / 2\) from mean position. Find phase difference in the two \(SHMs\).

364212 Two pendulums have time periods T and \(\frac{{5T}}{4}\).
They start S.H.M. at the same time from the mean position. What will be the phase difference between them after the bigger pendulum
complete one oscillation?

364213 The equation of S.H.M is \(y=a \sin (2 \pi n t+\alpha)\), then its phase at time \(t\) is

364214 Two simple harmonic motions exhibited by two particles \(A\) and \(B\) are given respectively by the following equations.
\({y_1} = a\sin \left[ {\omega t + \frac{\pi }{6}} \right]\)
\({y_2} = a\sin \left[ {\omega t + \frac{{3\pi }}{6}} \right]\)
Phase difference between them is

364210 The displacement-time equation of particle executing SHM is: \(x=A \sin (\omega t+\phi)\). At time \(t = 0\), position of the particle is \(x = A/2\) and it is moving along negative \(x\)-direction. Then the angle \(\phi\) can be:

364211 Two particles execute SHM with same amplitudes and same angular frequency on same straight line with same mean position. Given that during oscillation they cross each other in opposite direction when at a distance \(A / 2\) from mean position. Find phase difference in the two \(SHMs\).

364212 Two pendulums have time periods T and \(\frac{{5T}}{4}\).
They start S.H.M. at the same time from the mean position. What will be the phase difference between them after the bigger pendulum
complete one oscillation?

364213 The equation of S.H.M is \(y=a \sin (2 \pi n t+\alpha)\), then its phase at time \(t\) is

364214 Two simple harmonic motions exhibited by two particles \(A\) and \(B\) are given respectively by the following equations.
\({y_1} = a\sin \left[ {\omega t + \frac{\pi }{6}} \right]\)
\({y_2} = a\sin \left[ {\omega t + \frac{{3\pi }}{6}} \right]\)
Phase difference between them is

364210 The displacement-time equation of particle executing SHM is: \(x=A \sin (\omega t+\phi)\). At time \(t = 0\), position of the particle is \(x = A/2\) and it is moving along negative \(x\)-direction. Then the angle \(\phi\) can be:

364211 Two particles execute SHM with same amplitudes and same angular frequency on same straight line with same mean position. Given that during oscillation they cross each other in opposite direction when at a distance \(A / 2\) from mean position. Find phase difference in the two \(SHMs\).

364212 Two pendulums have time periods T and \(\frac{{5T}}{4}\).
They start S.H.M. at the same time from the mean position. What will be the phase difference between them after the bigger pendulum
complete one oscillation?

364213 The equation of S.H.M is \(y=a \sin (2 \pi n t+\alpha)\), then its phase at time \(t\) is

364214 Two simple harmonic motions exhibited by two particles \(A\) and \(B\) are given respectively by the following equations.
\({y_1} = a\sin \left[ {\omega t + \frac{\pi }{6}} \right]\)
\({y_2} = a\sin \left[ {\omega t + \frac{{3\pi }}{6}} \right]\)
Phase difference between them is

364210 The displacement-time equation of particle executing SHM is: \(x=A \sin (\omega t+\phi)\). At time \(t = 0\), position of the particle is \(x = A/2\) and it is moving along negative \(x\)-direction. Then the angle \(\phi\) can be:

364211 Two particles execute SHM with same amplitudes and same angular frequency on same straight line with same mean position. Given that during oscillation they cross each other in opposite direction when at a distance \(A / 2\) from mean position. Find phase difference in the two \(SHMs\).

364212 Two pendulums have time periods T and \(\frac{{5T}}{4}\).
They start S.H.M. at the same time from the mean position. What will be the phase difference between them after the bigger pendulum
complete one oscillation?

364213 The equation of S.H.M is \(y=a \sin (2 \pi n t+\alpha)\), then its phase at time \(t\) is

364214 Two simple harmonic motions exhibited by two particles \(A\) and \(B\) are given respectively by the following equations.
\({y_1} = a\sin \left[ {\omega t + \frac{\pi }{6}} \right]\)
\({y_2} = a\sin \left[ {\omega t + \frac{{3\pi }}{6}} \right]\)
Phase difference between them is