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Chapter 10: Problem 1339

A person standing in a stationary lift measures the periodic time of a simple pendulum inside the lift to be equal to \(\mathrm{T}\). Now, if the lift moves along the vertically upward direction with an acceleration of $(\mathrm{g} / 3)$, then the periodic time of the lift will now be \((\mathrm{A}) \sqrt{3} \mathrm{~T}\) (B) \(\sqrt{(3 / 2) \mathrm{T}}\) (C) \((\mathrm{T} / 3)\) (D) \((\mathrm{T} / \sqrt{3})\)

Short Answer

Expert verified
(B) \(\sqrt{(3 / 2) T}\)

Step by step solution

01

Determine the new effective acceleration due to gravity

The lift is accelerating vertically upward at \(\frac{g}{3}\), therefore the new effective acceleration due to gravity acting on the pendulum is \(g'\), where \(g' = g + \frac{g}{3}\). This is because the upward acceleration opposes the force of gravity, effectively increasing it.
02

Calculate the new periodic time T'

Now that we have the new effective acceleration due to gravity, we can use the formula for the periodic time of a simple pendulum to find the new periodic time T': \[T' = 2\pi\sqrt{\frac{L}{g'}}\]
03

Express T' in terms of T

We know that the original periodic time T is given by \(T = 2\pi\sqrt{\frac{L}{g}}\). We want to express T' in terms of T, so we need to eliminate the length L from the equation. Using the formula for T, solve for L: \[L = \frac{gT^2}{4\pi^2}\] Now substitute L in the equation for T': \[T' = 2\pi\sqrt{\frac{gT^2}{4\pi^2(g+\frac{g}{3})}}\]
04

Simplify the expression for T'

Now we simplify the expression for T': \[T' = 2\pi\sqrt{\frac{gT^2}{4\pi^2(\frac{4}{3}g)}}\] And further simplifying the expression: \[T' = \sqrt{\frac{3}{4}}T\] Now we compare the expression derived above with the given options. It matches the option: (B) \(\sqrt{(3 / 2) T}\) So the correct answer is option (B).

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