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Chapter 9: Problem 256

The length of a simple pendulum executing simple harmonic motion is increased by \(21 \%\). The percentage increase in the time period of the pendulum of increased length is (A) \(11 \%\) (B) \(21 \%\) (C) \(42 \%\) (D) \(10 \%\)

Short Answer

Expert verified
The percentage increase in the time period of the pendulum of increased length is approximately \(10 \%\) (Option D).

Step by step solution

01

Identify the formula for the time period of a simple pendulum

The time period (T) of a simple pendulum is given by the formula: \( T = 2\pi \sqrt{\cfrac{l}{g}} \) where "l" is the length of the pendulum and "g" is the acceleration due to gravity.
02

Calculate the new length of the pendulum

To find the new length after increasing by \(21\%\), we can represent the initial length as "l" and the new length as "L". Then we have: \( L = l + 0.21l = 1.21l \)
03

Calculate the new time period of the pendulum

Now, we'll use the formula for the time period to calculate the new time period (T') using the new length (L): \( T' = 2\pi \sqrt{\cfrac{L}{g}} = 2\pi \sqrt{\cfrac{1.21l}{g}} \)
04

Relate the two time periods

We'll take the ratio of the new time period (T') to the original time period (T): \( \cfrac{T'}{T} = \cfrac{2\pi \sqrt{\cfrac{1.21l}{g}}}{2\pi \sqrt{\cfrac{l}{g}}} = \sqrt{1.21} \)
05

Calculate the percentage increase in the time period

We'll find out what percentage increase corresponds to the ratio of the two time periods: Percentage increase \( = \left(\cfrac{T'}{T} - 1\right) \times 100 \% = \left(\sqrt{1.21} - 1\right) \times 100 \% \approx 10 \% \) So, the correct answer is (D) \(10 \%\).

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Most popular questions from this chapter

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