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A horizontal tree branch is directly above another horizontal tree branch. The elevation of the higher branch is \(9.65 \mathrm{~m}\) above the ground, and the elevation of the lower branch is \(5.99 \mathrm{~m}\) above the ground. Some children decide to use the two branches to hold a tire swing. One end of the tire swing's rope is tied to the higher tree branch so that the bottom of the tire swing is \(0.470 \mathrm{~m}\) above the ground. This swing is thus a restricted pendulum. Starting with the complete length of the rope at an initial angle of \(14.2^{\circ}\) with respect to the vertical, how long does it take a child of mass \(29.9 \mathrm{~kg}\) to complete one swing back and forth?

Short Answer

Expert verified
Answer: The time it takes for the child to complete one back-and-forth swing on the tire swing is approximately 6.05 seconds.

Step by step solution

01

Calculate the length of the rope

We are given 3 values regarding the heights. The height of the higher tree branch (9.65 m), the height of the lower tree branch (5.99 m), and the height of the tire swing above the ground (0.47) when at its lowest position. We can find the length of the rope (L) by subtracting 0.47 from the height of the higher tree branch: L = 9.65 m - 0.47 m = 9.18 m
02

Calculate the period of the pendulum

The period of a pendulum (the time it takes to complete one back-and-forth swing) can be calculated using the formula: T = 2π √(L/g) Where T is the period, L is the length of the rope (9.18 m in this case), and g is the acceleration due to gravity (approximately 9.81 m/s²). So, we can plug in the values: T = 2π √((9.18 m)/(9.81 m/s²))
03

Solve for the period T

Now it's time to solve for the period. When calculation with the given values, we find: T = 2π √((9.18 m)/(9.81 m/s²)) ≈ 6.05 s
04

Write down the final result

Now we have the time it takes for the child of mass 29.9 kg to complete one swing (back and forth) on the tire swing, which is approximately: T ≈ 6.05 s

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Simple Pendulum
Imagine a weight suspended from a fixed point with a string or rod that is allowed to swing freely back and forth under the influence of gravity. This is a simple pendulum. It's a classic example of an oscillatory system and has been instrumental in our understanding of timekeeping and physics in general.

When discussing a simple pendulum, there are several key factors to consider. Its length, defined as the distance from the pivot point to the center of mass of the pendulum bob, is the most critical variable because it directly affects the pendulum's period. Other elements include the bob's mass, although it surprisingly does not impact the period, and the acceleration due to gravity, which is a constant that affects all gravitational phenomena on Earth. The motion of a simple pendulum is predictable and periodic, making it a perfect candidate to delve into the concepts of harmonic motion and calculating the period of oscillation.
Period of a Pendulum
The period of a pendulum is the time it takes for the pendulum to swing from its initial position to the furthest point in its trajectory and back again to the start. It is solely determined by the length of the pendulum and the acceleration due to gravity, an insight first discovered by Galileo Galilei.

Period Calculation Example

Using the exercise provided as reference, the period (\(T\)) can be calculated with the formula \(T = 2\pi \sqrt{L/g}\), where \(L\) is the length of the pendulum, and \(g\) is the acceleration due to gravity. This formula is a key concept in understanding how pendulums behave. Even in the exercise where the pendulum (the tire swing) is set into motion at a slight angle, the period remains a constant for small angles, making the approximation valid and simplifying the swing into harmonic motion.
Harmonic Motion
Harmonic motion, or more specifically simple harmonic motion, is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. It's exemplified perfectly by the swinging of a simple pendulum, where gravity provides the restoring force.

What makes this concept critical for understanding pendulums is that the motion is predictable and regular. For small oscillations, a pendulum's motion can be approximated to simple harmonic motion, which allows us to use the formula for the period mentioned earlier effectively. This model works under the assumption that the amplitude is small and the pendulum doesn't lose energy over time, thus no air resistance or friction at the pivot is considered. Real-world pendulums, like the tire swing in our exercise, often experience dampened oscillations over time, but for the purpose of calculating a single period, the simple harmonic motion model serves as an excellent approximation.

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Most popular questions from this chapter

A spring is hanging from the ceiling with a mass attached to it. The mass is pulled downward, causing it to oscillate vertically with simple harmonic motion. Which of the following will decrease the frequency of the oscillation? a) adding a second, identical spring with one end attached to the mass and the other to the ceiling b) adding a second, identical spring with one end attached to the mass and the other to the floor c) increasing the mass d) adding both springs, as described in (a) and (b).

When the amplitude of the oscillation of a mass on a stretched string is increased, why doesn't the period of oscillation also increase?

Pendulum A has a bob of mass \(m\) hung from a string of length \(L\); pendulum \(\mathrm{B}\) is identical to \(\mathrm{A}\) except its bob has mass \(2 m .\) Compare the frequencies of small oscillations of the two pendulums.

A mass, \(M=1.60 \mathrm{~kg}\), is attached to a wall by a spring with \(k=578 \mathrm{~N} / \mathrm{m} .\) The mass slides on a frictionless floor. The spring and mass are immersed in a fluid with a damping constant of \(6.40 \mathrm{~kg} / \mathrm{s}\). A horizontal force, \(F(t)=F_{\mathrm{d}} \cos \left(\omega_{\mathrm{d}} t\right),\) where \(F_{\mathrm{d}}=52.0 \mathrm{~N},\) is applied to the mass through a knob, causing the mass to oscillate back and forth. Neglect the mass of the spring and of the knob and rod. At what frequency will the amplitude of the mass's oscillation be greatest, and what is the maximum amplitude? If the driving frequency is reduced slightly (but the driving amplitude remains the same), at what frequency will the amplitude of the mass's oscillation be half of the maximum amplitude?

A grandfather clock uses a pendulum and a weight. The pendulum has a period of \(2.00 \mathrm{~s},\) and the mass of the bob is \(250 . \mathrm{g} .\) The weight slowly falls, providing the energy to overcome the damping of the pendulum due to friction. The weight has a mass of \(1.00 \mathrm{~kg},\) and it moves down \(25.0 \mathrm{~cm}\) every day. Find \(Q\) for this clock. Assume that the amplitude of the oscillation of the pendulum is \(10.0^{\circ}\).

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