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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 = 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

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 = 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)

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Most popular questions from this chapter

A mass-spring system oscillates so that the position of the mass is described by \(x=-10 \cos (1.57 t),\) where \(x\) is in \(\mathrm{cm}\) when \(t\) is in seconds. Make a plot that has a dot for the position of the mass at $t=0, t=0.2 \mathrm{s}, t=0.4 \mathrm{s}, \ldots, t=4 \mathrm{s}$ The time interval between each dot should be 0.2 s. From your plot, tell where the mass is moving fastest and where slowest. How do you know?
A bungee jumper leaps from a bridge and undergoes a series of oscillations. Assume \(g=9.78 \mathrm{m} / \mathrm{s}^{2} .\) (a) If a \(60.0-\mathrm{kg}\) jumper uses a bungee cord that has an unstretched length of \(33.0 \mathrm{m}\) and she jumps from a height of \(50.0 \mathrm{m}\) above a river, coming to rest just a few centimeters above the water surface on the first downward descent, what is the period of the oscillations? Assume the bungee cord follows Hooke's law. (b) The next jumper in line has a mass of \(80.0 \mathrm{kg} .\) Should he jump using the same cord? Explain.
An ideal spring has a spring constant \(k=25 \mathrm{N} / \mathrm{m}\). The spring is suspended vertically. A 1.0 -kg body is attached to the unstretched spring and released. It then performs oscillations. (a) What is the magnitude of the acceleration of the body when the extension of the spring is a maximum? (b) What is the maximum extension of the spring?
An object of mass \(m\) is hung from the base of an ideal spring that is suspended from the ceiling. The spring has a spring constant \(k .\) The object is pulled down a distance \(D\) from equilibrium and released. Later, the same system is set oscillating by pulling the object down a distance \(2 D\) from equilibrium and then releasing it. (a) How do the period and frequency of oscillation change when the initial displacement is increased from \(D\) to $2 D ?$ (b) How does the total energy of oscillation change when the initial displacement is increased from \(D\) to \(2 D ?\) Give the answer as a numerical ratio. (c) The mass-spring system is set into oscillation a third time. This time the object is pulled down a distance of \(2 D\) and then given a push downward some more, so that it has an initial speed \(v_{i}\) downward. How do the period and frequency of oscillation compare to those you found in part (a)? (d) How does the total energy compare to when the object was released from rest at a displacement \(2 D ?\)
The period of oscillation of a spring-and-mass system is \(0.50 \mathrm{s}\) and the amplitude is \(5.0 \mathrm{cm} .\) What is the magnitude of the acceleration at the point of maximum extension of the spring?
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