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Chapter 10: Problem 87

A gibbon, hanging onto a horizontal tree branch with one arm, swings with a small amplitude. The gibbon's CM is 0.40 m from the branch and its rotational inertia divided by its mass is \(I / m=0.25 \mathrm{m}^{2} .\) Estimate the frequency of oscillation.

Short Answer

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#tag_title# Step 2: Calculate the period of oscillation#tag_content# Substitute the given values into the formula: T = 2蟺 鈭�(0.40 m / 9.81 m/s虏) T 鈮� 2蟺 鈭�(0.0408 s虏) T 鈮� 2蟺 脳 0.202 s T 鈮� 1.27 s The period of oscillation is approximately 1.27 seconds. #tag_title# Step 3: Find the frequency of oscillation#tag_content# Now that we have the period of oscillation, we can easily find the frequency of oscillation by taking the reciprocal of the period: Frequency (f) = 1 / T f 鈮� 1 / 1.27 s f 鈮� 0.79 Hz The frequency of oscillation of the gibbon hanging from the horizontal tree branch is approximately 0.79 Hz.

Step by step solution

01

Find the period of oscillation for a simple pendulum

In the case of a simple pendulum, the period of oscillation T can be expressed as: T = 2蟺 鈭�(L/g) where L is the distance from the pivot point to the center of mass, and g is the acceleration due to gravity (approximately 9.81 m/s虏). Here, L is given as 0.40 m. So, we can plug the values into the equation to obtain the period of oscillation. T = 2蟺 鈭�(0.40 m / 9.81 m/s虏)

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