Question

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

#tag_title# Step 2: Calculate the period of oscillation#tag_content# Substitute the given values into the formula: T = 2π √(0.40 m / 9.81 m/s²) T ≈ 2π √(0.0408 s²) T ≈ 2π × 0.202 s T ≈ 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 ≈ 1 / 1.27 s f ≈ 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

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 = 2π √(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/s²). Here, L is given as 0.40 m. So, we can plug the values into the equation to obtain the period of oscillation. T = 2π √(0.40 m / 9.81 m/s²)