Physics · Oscillations · NEET
Their time periods are different. Time period T = 2 pi root(L/g), so T depends on length L. A longer pendulum takes more time for one swing than a shorter one. Even if both start at the mean position moving the same way, after one swing the shorter one is already ahead. This small lead grows every swing until they are completely out of step, then keeps growing until the lead becomes exactly one full extra cycle and they line up again.
They are back in phase when both are at the mean position moving the same way at the same instant. That means n times T(short) = m times T(long), where n and m are the smallest whole numbers that make the two total times equal. Since T is proportional to root(L), take the ratio T(long)/T(short) = root(L_long/L_short), write it as a fraction m/n in lowest terms, and n = number of swings of the shorter pendulum, m = number of swings of the longer one.
The SHORTER pendulum. It has the smaller time period, so it swings faster and fits in more complete vibrations in the same total time. In the NEET 2022 problem the 100 cm (shorter) pendulum does 11 vibrations while the 121 cm (longer) pendulum does only 10. A common trap is to give the answer for the longer pendulum by mistake.
Because the lengths were chosen so root(L_long/L_short) is a nice fraction. Here root(121/100) = 11/10 exactly. NEET picks 121 and 100 (and similar perfect squares) on purpose so the square root gives a simple ratio. If the ratio were not clean you would get large or non-integer numbers, which NEET avoids for a one-minute question.
For the standard NEET question, yes, you look for them meeting at the mean position moving in the same direction, which is the true in-phase condition. They can cross paths at other points earlier, but that is not the same as being in phase. Being in phase means same displacement, same velocity direction, at the same time, which first repeats after the LCM of the two periods.
Two pendulums of lengths 121 cm and 100 cm start swinging in phase, both at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Set n times T(short) = m times T(long) for the smallest whole numbers n and m. Since T is proportional to root(L), compute the ratio root(L_long/L_short), write it as m/n in lowest terms, and n gives the number of swings of the shorter pendulum.
No. The time period of a simple pendulum, T = 2 pi root(L/g), does not depend on the mass of the bob. Only the lengths (and g) decide the periods, so mass never enters the in-phase calculation.
Then the ratio of periods will not be a simple fraction and the number of swings can be large or awkward. NEET almost always chooses lengths that are perfect squares (like 100, 121, 144) so root(L_long/L_short) is a neat fraction and the answer is a small whole number.
No. Two pendulums can cross the same point at different times or moving in opposite directions. Being in phase means same displacement AND same velocity direction at the same instant, which first repeats after the LCM of the two time periods.
