Dance of the pendulums

The least common multiple of the periods is 42 seconds; this is the first time $t > 0$ when all three pendulums have returned to their initial phase, after respectively 35, 20, and 14 full periods. However, this is not the correct answer because there are earlier moments at which the pendulums have the same phase.

The correct answer is $\boxed{14\ \text{seconds}}$ . At this moment, the number of periods completed by each pendulum is $\phi_1 = \frac{14}{1.2} = 11\tfrac23;\ \ \ \phi_2 = \frac{14}{2.1} = 6\tfrac23;\ \ \ \phi_3 = \frac{14}{3.0} = 4\tfrac23.$ Since each of the fractional parts is equal to $\phi* = 2/3$ , all pendulum will at this time have the same phase.

Generalized solution

Let $\tau = \gcd(T_1,\dots,T_N)$ be the greatest common divisor of all periods. Then $T_i = n_i\tau$ for integers $n_i$ .

Let $N = \text{lcm}(n_1,\dots,n_N)$ be the least common multiple of these integers. Then $T = N\tau$ is the first time at which the pendulums all have completed a whole number of periods and are therefore all in their initial phase. The fraction $\phi_i = \frac{T}{T_i} = \frac{N}{n_i}$ is the (integral) number of periods completed by the $i$ th pendulum.

Now consider the differences between the numbers of periods completed by two neighboring pendulums: $\Delta_i = \phi_{i+1} - \phi_i\ \ \ \ \ 1 \leq i < N$ Let $\delta = \gcd(\Delta_1,\dots,\Delta_{N-1})$ be the greatest common divisor of these differences. Let $t = \frac{N\tau}{\delta}.$ At this time, each pendulum has completed $\phi_i/\delta$ periods, which is not necessarily an integer. However, the differences between the $\phi_i/\delta$ are all integers, showing that the pendulums have the same phase.

Application to the problem at hand :

• $\tau = \gcd(1.2, 2.1, 3.0) = \SI{0.3}{\second}$ ;

• $n_1 = 4, n_2 = 7, n_3 = 10$ , so that $N = 140$ ;

• $T = 140\cdot 0.3 = \SI{42}{\second}$ ;

• $\phi_1 = 35, \phi_2 = 20, \phi_3 = 14$ ;

• $\Delta_1 = 15, \Delta_2 = 6$ , so that $\delta = 3$ ;

• $t = 42/3 = \SI{14}{\second}$ .

It is mentioned in the problem that the amplitude of oscillation is small. Therefore the pendulums can be considered in simple harmonic motion(SHM) according to what I learnt in school. The equation of SHM is....... Y=A.cos(2πt/T) ....Where T is the time period. But here amplitudes are same (A). So the relative positions depend upon cos(2πt/T). Now we have to find first instance ,t>0 when cos(2πt/1.2) , cos(2πt/2.1), cos(2πt/3) are equal. This happens first when t=14 sec.

LCM(A, B,C)=LCM (1.2, 2.1, 3) =.3 * LCM(4, 7, 10)=.3 *140=42.
So after 42 sec. they meet at starting point after 35, 20, 14 cycles.
Let us see if they meet any time before.
$In~ 42/2=21sec,~ A:-~~17\frac 1 2 cycles.~~~~~B:-~~10 cycles.~~~~~C:-~~7cycles.~~~~~A ~out ~of~ phase. \\ In~ 42/3=14sec,~ A:-~~\dfrac {14}{1.2} =11\frac 2 3 cycles.~~~~~B:-~~\dfrac {14}{2.1} =6\frac 2 3 cycles.~C:-~~~~\dfrac {14}{3} =4\frac 2 3.\\ All~three~have~the~same~fraction~\implies~they~are~in~phase.\\ They~are~in~phase~when~2/3-1/2=\frac 1 6 ^{th}~the~return~swing.\\ 42~divide~by~any~bigger~integer~does ~not~produce ~same~fraction.\\ So~after~{\large \color{#D61F06}{14 ~sec} },~ all ~are~in~phase ~for~the~first~time~after~start.$