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Chapter 10: Problem 58

A bob of mass \(m\) is suspended from a string of length \(L\) forming a pendulum. The period of this pendulum is \(2.0 \mathrm{s}\) If the pendulum bob is replaced with one of mass \(\frac{1}{3} m\) and the length of the pendulum is increased to \(2 L\), what is the period of oscillation?

Short Answer

Expert verified
Answer: The period of oscillation for the pendulum with the new length and mass is approximately \(2.8314 \mathrm{s}\).

Step by step solution

01

Recall the formula for the period of a pendulum

The formula for the period of a simple 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 (approximately \(9.81 \mathrm{m/s^2}\)). Note that the mass of the pendulum bob does not affect the period.
02

Calculate the initial value of \(L\)

We are given the period of the initial pendulum (\(T_1 = 2.0 \mathrm{s}\)). We can use this to find the length of the pendulum, \(L\). Rearrange the formula to make \(L\) the subject: \(L = \frac{gT^2}{4\pi^2}\). Now, plug in the values: \(L = \frac{9.81 \cdot 2^2}{4 \cdot \pi^2} \approx 0.9934 \mathrm{m}\).
03

Calculate the new length and mass

We are told that the length of the pendulum is increased to \(2L\), and the mass of the bob is changed to \(\frac{1}{3} m\). However, since the mass does not affect the period, we don't need to consider it in our calculation. The new length is \(L_2 = 2L = 2(0.9934) = 1.9868 \mathrm{m}\).
04

Calculate the new period

Now, we can find the new period \(T_2\) using the new length \(L_2\). Substitute the new length into the formula: \(T_2 = 2\pi\sqrt{\frac{L_2}{g}} = 2\pi\sqrt{\frac{1.9868}{9.81}} \approx 2.8314 \mathrm{s}\).
05

Answer

The period of oscillation for the pendulum with the new length and mass is approximately \(2.8314 \mathrm{s}\).

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