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Chapter 14: Problem 10

14.10 A pendulum is suspended from the ceiling of an elevator. When the elevator is at rest, the period of the pendulum is \(T\). The elevator accelerates upward, and the period of the pendulum is then a) still T. b) less than \(T_{-}\) c) greater than \(T\).

Short Answer

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Question: When an elevator accelerates upward, the period of a pendulum suspended from the ceiling of the elevator becomes: a) equal to the original period \(T\) b) less than the original period \(T\) c) greater than the original period \(T\) Answer: b) less than the original period \(T\)

Step by step solution

01

State the formula for the period of a pendulum

The formula for the period of a pendulum is given by: \[T = 2\pi\sqrt{\frac{l}{g}}\] where \(T\) is the period, \(l\) is the length of the pendulum, and \(g\) is the acceleration due to gravity.
02

Analyze the effect of the elevator's upward acceleration on the pendulum

When the elevator accelerates upward, it creates an additional force acting on the pendulum, in the upward direction. This force increases the effective gravity acting on the pendulum. Let's represent the upward acceleration as \(a\). Since the acceleration is acting in the same direction as gravity, the effective gravity acting on the pendulum can be written as: \[g' = g + a\]
03

Find the period of the pendulum with the new effective gravity

Since the new effective gravity is \(g'\), the formula for the period of the pendulum when the elevator is accelerating upward can be written as: \[T' = 2\pi\sqrt{\frac{l}{g'}}\] Substitute the value of \(g'\) from step 2: \[T' = 2\pi\sqrt{\frac{l}{g + a}}\]
04

Compare the new period \(T'\) with the original period \(T\)

We can compare the two periods by finding the ratio between them: \[\frac{T'}{T} = \frac{2\pi\sqrt{\frac{l}{g + a}}}{2\pi\sqrt{\frac{l}{g}}}\] The \(2\pi\sqrt{\frac{l}{g}}\) terms cancel out: \[\frac{T'}{T} = \sqrt{\frac{g}{g+a}}\] Since \(g > 0\) and \(a > 0\), the denominator \((g+a)\) will always be greater than \(g\). Therefore, as the denominator increases while the numerator remains constant, the ratio \(\frac{g}{g + a}\) will be less than 1. Thus, the period \(T'\) of the pendulum when the elevator is accelerating upward will be less than the original period \(T\).
05

State the answer

Based on our analysis, the period of the pendulum \(T'\) when the elevator is accelerating upward will be less than the original period \(T\). Hence, the correct answer is: (b) less than \(T\).

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Most popular questions from this chapter

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