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Chapter 10: Problem 1372

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

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

Step by step solution

01

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 \(9.81 m/s^2\)).
02

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: \[T_2 = 2\pi\sqrt{\frac{1.21}{9.81}}\]
03

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.
04

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.

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