Pendulum Vibrations In Phase Analysis

This problem explores the concept of synchronization between two pendulums of different lengths. We are given two pendulums, one with length $L_1 = 169$ cm and another with length $L_2 = 144$ cm. They start their motion together, in phase, from the mean position. The goal is to determine the minimum number of vibrations the shorter pendulum must complete before both pendulums are simultaneously in phase again at their mean position.

Understanding Pendulum Time Period

The time period ($T$) of a simple pendulum represents the time it takes to complete one full oscillation (back and forth). This period is primarily determined by the pendulum's length ($L$) and the local acceleration due to gravity ($g$). The formula governing the time period is:

$T = 2\pi \sqrt{\frac{L}{g}}$

In this scenario, both pendulums are subject to the same gravitational acceleration ($g$). Therefore, the time period of each pendulum is directly proportional to the square root of its length ($T \propto \sqrt{L}$).

Calculating Time Period Ratio for Given Lengths

Let $T_1$ denote the time period of the first pendulum with length $L_1 = 169$ cm, and $T_2$ denote the time period of the second pendulum with length $L_2 = 144$ cm.

Using the formula, we can express their time periods as:

$T_1 = 2\pi \sqrt{\frac{169}{g}}$

$T_2 = 2\pi \sqrt{\frac{144}{g}}$

To understand how their periods relate, we calculate the ratio $\frac{T_1}{T_2}$:

$ \frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{169}{g}}}{2\pi \sqrt{\frac{144}{g}}} $

Simplifying this expression gives:

$ \frac{T_1}{T_2} = \sqrt{\frac{169}{144}} $

Taking the square root of the numerator and denominator:

$ \frac{T_1}{T_2} = \frac{\sqrt{169}}{\sqrt{144}} $

$ \frac{T_1}{T_2} = \frac{13}{12} $

This ratio indicates that $T_1$ is slightly longer than $T_2$. Rearranging this, we get the relationship $12 T_1 = 13 T_2$.

Condition for Pendulums Being In Phase

The problem states that the pendulums start vibrating in phase from the mean position. For them to be in the same phase at the mean position again, both must have completed an integer number of full oscillations at the exact same moment in time. Let $n_1$ be the number of vibrations completed by the first pendulum ($L_1$) and $n_2$ be the number of vibrations completed by the second pendulum ($L_2$) when this condition is met.

The total time elapsed for $n_1$ vibrations of the first pendulum is $t = n_1 T_1$.

Similarly, the total time elapsed for $n_2$ vibrations of the second pendulum is $t = n_2 T_2$.

Since they must complete these vibrations in the same amount of time to be in phase again:

$ n_1 T_1 = n_2 T_2 $

Calculating Minimum Vibrations

We can use the ratio of the time periods ($\frac{T_1}{T_2} = \frac{13}{12}$) to find the relationship between the number of vibrations $n_1$ and $n_2$. We know $T_1 = \frac{13}{12} T_2$. Substituting this into the phase synchronization equation:

$ n_1 \left( \frac{13}{12} T_2 \right) = n_2 T_2 $

We can divide both sides by $T_2$ (as $T_2$ is non-zero):

$ \frac{13 n_1}{12} = n_2 $

Multiplying both sides by 12 yields:

$ 13 n_1 = 12 n_2 $

We need to find the smallest positive integers $n_1$ and $n_2$ that satisfy this equation. Since 13 and 12 share no common factors other than 1 (they are coprime), the smallest integer solution is obtained by setting:

• $n_1 = 12$
• $n_2 = 13$

This means that when the first pendulum (169 cm) completes 12 vibrations, the second pendulum (144 cm) simultaneously completes 13 vibrations. At this point, they will both be back at the mean position, vibrating in the same phase.

Identifying Shorter Pendulum's Vibrations

The question specifically asks for the minimum number of vibrations of the shorter pendulum. Comparing the lengths, $L_2 = 144$ cm is shorter than $L_1 = 169$ cm.

The number of vibrations corresponding to the shorter pendulum ($L_2$) is represented by $n_2$. Our calculation shows that the minimum integer value for $n_2$ is 13.

Conclusion

Thus, the minimum number of vibrations the shorter pendulum (144 cm) must complete for both pendulums to be in phase at the mean position again is 13.