A simple pendulum (whose length is less than that of a second's pendulum) and a second's pendulum start swinging in phase. They again swing in phase after an interval of $18$ seconds from the start. The period of the simple pendulum is
(A) $0.9$ sec
(B) $1.8$ sec
(C) $2.7$ sec
(D) $3.6$ sec
I was given a formula for such questions:
$$T = \frac {T_1 T_2} {T_1-T_2} \qquad (T_1>T_2)$$
where $T_1$ and $T_2$ are the time periods of the individual pendulums, and $T$ is the time after which they are in phase again.
I took $T_1$ as the seconds pendulum, i.e., $T_1=2$ seconds.
Using the formula, I got $T_2=1.8$ sec, which makes sense; the timestamps for each oscillation are:
$$1.8\ \ 3.6\ \ 5.4\ \ 7.2\ \ 9.0\ \ 10.8\ \ 12.6\ \ 14.4\ \ 16.8\ \ 18.0$$
seconds for simple pendulum, and $2, 4, 6, 8, 10, 12, 14, 16, 18$ seconds for seconds pendulum. None of these overlap, so if $T_2=1.8$, the pendulums swing in phase after intervals of $18$ seconds.
However, I also tried option A, and got the timestamps as:
$$0.9\ \ 1.8\ \ 2.7\ \ 3.6\ \ 4.5\ \ 5.4\ \ 6.3\ \ 7.2\ \ 8.1\ \ 9.0\ \ 9.9\ \ 10.8\ \ 11.7\ \ 12.6\ \ 13.5\ \ 14.4\ \ 15.3\ \ 16.2\ \ 17.1\ \ 18$$
seconds for simple pendulum, and $2, 4, 6, 10, 12, 14, 16, 18$ seconds for seconds pendulum. Again, none of these overlap, so if $T_2=0.9$ seconds also, the pendulums swing in phase after intervals of $18$ seconds.
According to the answer key, the answer is only B. Is A also correct, or am I missing something?
Best Answer
The key point that's overlooked in the timestamp-counting method is that having the pendulums be in sync at the end of complete periods is not the only way for them to be in phase - they can also happen to be in phase in the middle of a period. In particular, for this example, note that after $\frac{18}{11}$ seconds, the $0.9$-second-period pendulum and the $2$-second-period pendulum will be $\frac{9}{11}$ of the way through a period (try dividing $\frac{18}{11}$ seconds by each of their periods and verify for yourself). By looking only at timestamps of complete periods, the timestamp-counting method misses out this point (earlier than $18$ seconds) where they came back in phase.
I'd highlight that this means care is needed to derive the $\frac{T_1 T_2}{T_1 - T_2}$ formula - for example, it's not enough to just solve for the times when the pendulums have the same (angular) position, because there are many earlier times where this happens, but requiring that the pendulums are in phase is a much stronger condition. Also, one has to explicitly use the fact that we are interested in the first time they are back in phase, because it's true that the $0.9$-second-period pendulum and the $2$-second-period pendulum are in phase after $18$ seconds - the tricky thing is that there was an earlier time where they were already in phase. Basically, the correct way to derive that formula would be to say we are solving for the earliest time $t$ such that the difference between $t/T_1$ and $t/T_2$ is an integer.
For the explicit derivation: the pendulums are in phase at time $t$ if and only if $t/T_1 - t/T_2 = n$ for some integer $n$. Solving for $t$ yields
$$t = n \frac{T_1 T_2}{T_1 - T_2},$$
and hence we see that they are in phase whenever $t$ is an integer multiple of $\frac{T_1 T_2}{T_1 - T_2}$ (to restrict to positive $t$, take $T_1 > T_2$ and $n>0$ without loss of generality). In particular, the first positive $t$ at which this occurs is clearly when $n=1$, i.e. $t=\frac{T_1 T_2}{T_1 - T_2}$ as claimed.