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Chapter 10: Problem 61

A pendulum of length \(L_{1}\) has a period \(T_{1}=0.950 \mathrm{s}\). The length of the pendulum is adjusted to a new value \(L_{2}\) such that $T_{2}=1.00 \mathrm{s} .\( What is the ratio \)L_{2} / L_{1} ?$

Short Answer

Expert verified
Answer: The ratio of the adjusted length to the initial length of the simple pendulum is approximately 1.11.

Step by step solution

01

Recalling the formula for the period of a simple pendulum

The formula for the period \(T\) of a simple pendulum of length \(L\) and small oscillation amplitude is given by: $$ T = 2 \pi \sqrt{\frac{L}{g}} $$ where \(g\) is the acceleration due to gravity (\(\approx 9.81 \mathrm{m/s^2}\)).
02

Applying the formula to the initial length and period

Given the initial period \(T_1 = 0.950 \ \mathrm{s}\) and length \(L_1\), we apply the formula for the period of a simple pendulum: $$ T_1 = 2 \pi \sqrt{\frac{L_1}{g}} $$
03

Applying the formula to the adjusted length and period

Given the final period \(T_2 = 1.00 \ \mathrm{s}\) and the new length \(L_2\), we apply the formula for the period of a simple pendulum: $$ T_2 = 2 \pi \sqrt{\frac{L_2}{g}} $$
04

Writing the ratio of the two periods

Let's divide the equation for \(T_2\) by the equation for \(T_1\), which gives us the ratio of the two periods: $$ \frac{T_2}{T_1} = \frac{2 \pi \sqrt{\frac{L_2}{g}}}{2 \pi \sqrt{\frac{L_1}{g}}} $$
05

Simplifying the expression

Cancelling the common terms and cross-multiplying, we have: $$ \frac{T_2}{T_1} = \frac{\sqrt{L_2}}{\sqrt{L_1}} $$ Now, square both sides of the equation to get rid of the square roots: $$ \left(\frac{T_2}{T_1}\right)^2 = \frac{L_2}{L_1} $$
06

Solving for the ratio \(L_2 / L_1\)

The problem asks for the ratio \(L_2 / L_1\). Plug the given values of \(T_1\) and \(T_2\) into the equation above: $$ \frac{L_2}{L_1} = \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{1.00 \ \mathrm{s}}{0.950 \ \mathrm{s}}\right)^2 $$ Now, calculate the ratio: $$ \frac{L_2}{L_1} \approx 1.11 $$ So the ratio \(L_2 / L_1\) is approximately \(1.11\).

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