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Chapter 9: Problem 156

Two pendulums of same amplitude but time period \(3 \mathrm{~s}\) and \(7 \mathrm{~s}\) start oscillating simultaneously from two opposite extreme positions. After how much time they will be in phase (A) \(\frac{21}{8} \mathrm{~s}\) (B) \(\frac{21}{4} \mathrm{~s}\) (C) \(\frac{21}{2} \mathrm{~s}\) (D) \(\frac{21}{10} \mathrm{~s}\)

Short Answer

Expert verified
The time it takes for both pendulums to be in phase again is found by calculating the Least Common Multiple (LCM) of their half periods, which is \(\frac{21}{2} \mathrm{~s}\). Therefore, the correct answer is (C) \(\frac{21}{2} \mathrm{~s}\).

Step by step solution

01

Identify the time periods of both pendulums

The first pendulum has a time period of \(3 \mathrm{~s}\), and the second pendulum has a time period of \(7 \mathrm{~s}\).
02

Calculate the time for both pendulums to reach their opposite extreme position

To reach their opposite extreme positions, both pendulums must complete half of their oscillations. Let's calculate the time it takes for each pendulum to complete half of an oscillation. For the first pendulum with a time period of \(3 \mathrm{~s}\), half the time period is: \(\frac{1}{2} \times 3 \mathrm{~s} = \frac{3}{2} \mathrm{~s}\) For the second pendulum with a time period of \(7 \mathrm{~s}\), half the time period is: \(\frac{1}{2} \times 7 \mathrm{~s} = \frac{7}{2} \mathrm{~s}\)
03

Find the Least Common Multiple (LCM) of the half periods of both pendulums

To find the time it takes for both pendulums to be in phase again, we must find the least common multiple (LCM) of their half periods. LCM of \(\frac{3}{2} \mathrm{~s}\) and \(\frac{7}{2} \mathrm{~s}\): \( \mathrm{LCM}(\frac{3}{2}, \frac{7}{2}) = \frac{\mathrm{LCM}(3, 7)}{2} \) Since 3 and 7 are prime numbers, their LCM is simply their product: \( \frac{\mathrm{LCM}(3, 7)}{2} = \frac{3 \times 7}{2} = \frac{21}{2} \mathrm{~s} \)
04

Find the correct answer choice

The time it takes for both pendulums to be in phase again is \(\frac{21}{2} \mathrm{~s}\). Thus, the correct answer is: (C) \(\frac{21}{2} \mathrm{~s}\)

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