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Two pendulums with different frequencies released at the same time, when will these two pendulum be in phase?

From what I know, the period of pendulum at small displacement is not affected by its amplitude, so I tried to use the period formula $$T=2\pi \sqrt{\frac{l}{g}}$$ and substitute $l$, $g_{1}$ and $g_{2}$

But I am not sure how to proceed

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  • $\begingroup$ By "in phase," do you mean when the pendulums will be at the same angular displacement? I'm not sure if two oscillators are different frequencies can be "in phase." $\endgroup$
    – DanDan面
    Commented Sep 20, 2021 at 1:34
  • $\begingroup$ Yes, same angular displacement at the same time. Is it not possible? $\endgroup$
    – Sirou Ewei
    Commented Sep 20, 2021 at 1:42

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I've not heard people talk about two oscillators at different frequencies being "in phase." Instead, I will present a solution for time instants when the two oscillators will be at the same phase in their oscillation.

WLOG, let $\theta_1(0)=\theta_2(0)=0$. Define $\omega_i\equiv\sqrt{\frac{g_i}{l}}$ for $i\in\{1,2\}$; i.e. the angular frequency of the two oscillators.

Then, we have $\theta_1(t)=A_1\sin(\omega_1t)$ and $\theta_2(t)=A_2\sin(\omega_2t)$. To find when the two oscillators coincide in phase, we find solutions of $t$ to the following equation:

$$\sin(\omega_1t)=\sin(\omega_2t)$$

It's easiest to visualize solutions for this equation by graphing $\sin(x)=\sin(y)$. Solutions to this happen when $(\omega_1-\omega_2)t=2\pi i$ or $(\omega_1+\omega_2)t=\pi+2\pi i$, for $i\in\mathbb{Z}$.

Rearranged for $t$, we have $t=\frac{2\pi i}{\omega_1-\omega_2}$ and $t=\frac{\pi+2\pi i}{\omega_1+\omega_2}$.

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  • $\begingroup$ Could you explain how the last part with imaginary number comes in? If $sin(\omega_{1}t) = sin(\omega_{2}t)$, then by trig identity, wouldn't that require $\omega_{1}=\omega_{2}$ to be true? $\endgroup$
    – Sirou Ewei
    Commented Sep 20, 2021 at 2:38
  • $\begingroup$ I'm using $i$ simply for enumeration. Here, $i$ represents any integer. $\endgroup$
    – DanDan面
    Commented Sep 20, 2021 at 2:40
  • $\begingroup$ So $2\pi i$ is just a multiple of $2\pi$ ? $\endgroup$
    – Sirou Ewei
    Commented Sep 20, 2021 at 2:45
  • $\begingroup$ Yes, it can be $...,-2\pi,0,2\pi,4\pi,...$ $\endgroup$
    – DanDan面
    Commented Sep 20, 2021 at 2:48
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Let pendulums $P_1$ and $P_2$ have time period $T_1$ and $T_2$ and initial phase $\phi_1$ and $\phi_2$ respectively. For them to have the same phase, $$\sin\left(\frac{2\pi}{T_1}t+\phi_1\right) = \sin\left(\frac{2\pi}{T_2}t+\phi_2\right)$$$$\implies \frac{2\pi}{T_1}t+\phi_1 = n\pi + (-1)^n\left(\frac{2\pi}{T_2}t+\phi_2\right), n\in\mathbb Z$$From here, you can find solutions for $t$ for even and odd cases of $n$. These solutions of $t$ will be the instants when the two pendulums are in the same position on their phasor diagrams.

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  • $\begingroup$ What are the intermediate steps? Do you multiply both side by $\arcsin$ ? $\endgroup$
    – Sirou Ewei
    Commented Sep 20, 2021 at 3:10
  • $\begingroup$ This is a solution of trigonometric equation of the type sinx = siny. You can solve this equation by subtracting siny on both sides and write it as $2cos(\frac{x+y}{2})sin(\frac{x-y}{2}) = 0$. You will get $x=(2n+1)\pi - y$ and $x=2n\pi +y$. I hope you can find the combined solution set. $\endgroup$
    – Reet Jaiswal
    Commented Sep 20, 2021 at 3:34
  • $\begingroup$ Note that OP says the two pendulums are released at the same time. Consideration of $\phi_1$ and $\phi_2$ is unnecessary in this case $\endgroup$
    – DanDan面
    Commented Sep 20, 2021 at 5:13
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    $\begingroup$ @DanDan0101 I suppose you are right with the use of the word 'released' but I also took into account the cases where the bob was given some tangential velocity. $\endgroup$
    – Reet Jaiswal
    Commented Sep 20, 2021 at 5:34
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The general solution is:

Pendulum 1

$$\theta_1(t)=A_1\,\sin(\omega_1\,t)+B_1\,\cos(\omega_1\,t)$$

Pendulum 2

$$\theta_2(t)=A_2\,\sin(\omega_2\,t)+B_2\,\cos(\omega_2\,t)$$

obtain

$$\theta_1(t)-\theta_2(t)=A_1\,\sin(\omega_1\,t)-A_2\,\sin(\omega_2\,t)+ B_1\,\cos(\omega_1\,t)-B_2\,\cos(\omega_2\,t)\tag 1$$

and with:

$$\sin(x)-\sin(y)=2\,\sin \left( \frac{x-y}{2} \right) \cos \left( \frac{x+y}{2} \right)\\ \cos(x)-\cos(y)=-2\,\sin \left( \frac{x+y}{2} \right) \sin \left( \frac{x-y}{2} \right) $$

Eq. (1)

$$\theta_1(t)-\theta_2(t)=2\,A_1\,A_2\,\sin \left( \frac{t\,(\omega_1-\omega_2)}{2} \right) \cos \left( \frac{t\,(\omega_1+\omega_2)}{2} \right)\\ -2\,B_1\,B_2\,\sin \left( \frac{t\,(\omega_1+\omega_2)}{2} \right) \sin \left( \frac{t\,(\omega_1-\omega_2)}{2} \right)$$

thus

$\theta_1(t)=\theta_2(t)~$ only if $~\omega_1=\omega_2~$

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