Question

Two simple pendulums having lengths \(144 \mathrm{~cm}\) and \(121 \mathrm{~cm}\) starts executing oscillations. At some time, both bobs of the pendulum are at the equilibrium positions and in same phase. After how many oscillations of the shorter pendulum will both the bob's pass through the equilibrium position and will have same phase ? (A) 11 (B) 12 (C) 21 (D) 20

Short answer

The shorter pendulum will pass through the equilibrium position and have the same phase after \(132\) oscillations.

Step-by-step solution

Use the formula for the period of a simple pendulum.

For a simple pendulum, the period \(T\) is given by the formula: \[T = 2\pi\sqrt{\frac{L}{g}}\] where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity (approximately \(9.81 m/s^2\)).

Calculate the periods of the two pendulums.

Using the given lengths for the pendulums (\(144 \mathrm{~cm}\) and \(121 \mathrm{~cm}\)), we convert them to meters and find their periods. For the pendulum with length \(144 \mathrm{~cm}\) (1.44 meters), the period is: \[T_1 = 2\pi\sqrt{\frac{1.44}{9.81}}\] Similarly, for the pendulum with length \(121 \mathrm{~cm}\) (1.21 meters), the period is: \[T_2 = 2\pi\sqrt{\frac{1.21}{9.81}}\]

Calculate the ratio \(T_1/T_2\).

Calculate the ratio of the periods of the two pendulums to find the relationship between their oscillation rates. \[\frac{T_1}{T_2} = \frac{2\pi\sqrt{\frac{1.44}{9.81}}}{2\pi\sqrt{\frac{1.21}{9.81}}}\] Simplify, cancelling out the \(2\pi\) and \(9.81\): \[\frac{T_1}{T_2} = \frac{\sqrt{1.44}}{\sqrt{1.21}}\] Determine the square roots, and find the ratio: \[\frac{T_1}{T_2} = \frac{\sqrt{144}}{\sqrt{121}} = \frac{12}{11}\] This means that the first pendulum will complete 12 oscillations in the same time the second pendulum completes 11 oscillations.

Find the number of oscillations of the shorter pendulum.

Since it is given that the ratio of the oscillations is 12:11, it only takes the least common multiple (LCM) of 12 and 11 to find the number of oscillations for the shorter pendulum. LCM(12, 11) = 12 * 11 = 132 So the shorter pendulum passes through the equilibrium position and has the same phase after 132 oscillations. The correct answer is (C) 21.

Key concepts

Oscillation

An oscillation is a repetitive movement between two points. In the context of a pendulum, this means swinging back and forth between two extremes around a central point, known as the equilibrium position. One full oscillation includes the movement from the starting position to the opposite extreme and back again.

Understanding oscillations is crucial when studying pendulums because the regular and predictable nature of the oscillation provides a basis for calculating various properties like the period. Each pass through the equilibrium position marks a half cycle, and counting these gives us an idea of how many complete swings have occurred.

Period of a Pendulum

The period of a pendulum is the time it takes to complete one full oscillation. For a simple pendulum, this is determined by the formula:
\[T = 2\pi\sqrt{\frac{L}{g}}\]
where

• \(L\) is the length of the pendulum,
• \(g\) is the acceleration due to gravity (approximately 9.81 m/s²).

These factors make
- the period dependent only on the length of the pendulum and gravity.
This characteristic means that the period of a pendulum does not change with the amplitude of the swing or with the mass of the pendulum bob.
For instance, increasing the pendulum's length will increase the period, resulting in a slower swing, while reducing the length will decrease the period, making it oscillate faster.

Equilibrium Position

The equilibrium position of a pendulum is the central point or line around which the pendulum oscillates. This position is also sometimes referred to as the middle or rest position.
When a pendulum is not swinging, it naturally hangs downward due to gravity, aligning with the vertical equilibrium position.
When the pendulum swings, it moves away from and returns to this central position. Each time it passes through this point, it is momentarily at rest before accelerating in the opposite direction. Understanding the equilibrium position is valuable because the time it takes to pass through it marks significant moments in each oscillation cycle.

Least Common Multiple (LCM)

The least common multiple (LCM) is the smallest number that is a multiple of two or more integers. In problems involving oscillations, especially when considering when two oscillating systems will be synchronized again, finding the LCM can provide the answer.
For instance, if two pendulums have oscillation ratios of 12:11, as in the exercise, calculating their LCM indicates how many complete cycles of the shorter pendulum are needed for both to be back in the same phase.

• Computing the LCM of 12 and 11 involves determining the smallest common multiple of these two numbers, which is 132 in this example.

Thus, after 132 oscillations of the shorter pendulum, both pendulums meet again in the same phase at the equilibrium position.