Two pendulums have time-periods T and 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 has completed one oscillation? (in degree)

Question

Two pendulums have time-periods T and 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 has completed one oscillation? (in degree) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

When bigger pendulum completes one oscillation, smaller Pendulum completes (5/4) oscillation, i.e. difference of (1/4) oscillation. Hence φ=90°.

Q. Two pendulums have time-periods $T$ and $5 T /4$ . 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 has completed one oscillation? (in degree)

When bigger pendulum completes one oscillation, smaller Pendulum completes $\frac{5}{4}$ oscillation, i.e. difference of $\frac{1}{4}$ oscillation. Hence $ϕ = 9 0^{\circ}$ .

Q. Two pendulums have time-periods T and 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 has completed one oscillation? (in degree)

When bigger pendulum completes one oscillation, smaller Pendulum completes 45​ oscillation, i.e. difference of 41​ oscillation. Hence ϕ=90∘.

Q. Two pendulums have time-periods $T$ and $5 T /4$ . 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 has completed one oscillation? (in degree)

When bigger pendulum completes one oscillation, smaller Pendulum completes $\frac{5}{4}$ oscillation, i.e. difference of $\frac{1}{4}$ oscillation. Hence $ϕ = 9 0^{\circ}$ .