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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 \(0.47 \mathrm{~m}\) above the ground. This swing is thus a restricted pendulum. Start. ing 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 full swing back and forth is approximately 3.818 seconds.

Step by step solution

01

Calculate the Length of the Rope

The first step is to calculate the length of the rope. We are given that the higher branch is at 9.65 m and the tire swing bottom is 0.47 m above the ground. To do this, subtract the elevations of the bottom of the tire swing from the top branch: Length of the rope = elevation of higher branch - elevation of tire swing above ground L = 9.65 - 0.47 L = \(\boxed{9.18 \mathrm{~m}}\)
02

Calculate the Vertical Distance

Next, we'll calculate the vertical distance between the highest point of the swing and the lowest point. We know that the angle between the rope and the vertical is \(14.2^\circ\), so we can use the sine function to find the vertical distance (h): h = L * sin(angle) h = 9.18 * sin(14.2) h ≈ \(\boxed{2.27 \mathrm{~m}}\)
03

Calculate the Effective Length of the Pendulum

Now, we need to calculate the effective length of the pendulum, which is the vertical distance between the two branches: Effective length, Leff = elevation of higher branch - elevation of lower branch Leff = 9.65 - 5.99 Leff = \(\boxed{3.66 \mathrm{~m}}\)
04

Calculate the Period of the Pendulum

Now that we have the effective length of the pendulum, we can use the formula for the period of a pendulum to calculate the time it takes for one full swing back and forth: T = 2π * √ (Leff / g), where g is the acceleration due to gravity (approximately 9.81 m/s²) T = 2π * √ (3.66/9.81) T ≈ \(\boxed{3.818 \mathrm{~s}}\) The time it takes for the child to complete one swing back and forth is approximately 3.818 seconds.

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Key Concepts

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

Simple Pendulum
Let's explore the concept of a simple pendulum, which is a vital part of understanding pendulum problems. A simple pendulum consists of a weight, also known as a bob, attached to a string or rope that swings from a pivot point. The swing occurs in a plane under the influence of gravity alone, making it an ideal system for studying motion.
In our problem, the rope from the tire attached to the higher tree branch acts as the pendulum's length, with the tire forming the pendulum's bob. In essence, the simple pendulum is driven by the gravitational force acting on its mass as it swings back and forth. This system showcases fundamentals of periodic motion, which are essential to calculate, using trigonometry and other physics principles, how the pendulum behaves during its motion.
Oscillation Period
The period of oscillation is a crucial element in pendulum motion. It refers to the time it takes for a pendulum to complete one full cycle of swinging back and forth. This single cycle is called an oscillation. The period of oscillation is influenced by the pendulum's length and the gravitational force acting upon it.
To determine the oscillation period in our problem, we used the formula for the period of a simple pendulum: \[T = 2\pi \sqrt{\frac{L_{ ext{eff}}}{g}}\] Here, \(L_{\text{eff}}\) is the effective length of the pendulum and \(g\) equals the gravitational acceleration (approximately 9.81 m/s²). This formula illustrates that the period is directly proportional to the square root of the pendulum's length. Therefore, any increase in length results in a longer period, and vice versa. Understanding the oscillation period is integral to predicting the pendulum's motion over time and determining how long it takes for our tire swing to complete its cycle.
Trigonometry in Physics
Trigonometry often plays an essential role in physics, especially in analyzing systems involving angles and periodic motion like pendulums. It helps determine various components of motion, such as the height reached or the potential energy during the swing.
In our exercise, trigonometry was used to find the vertical distance the tire swing travels. Given that the angle between the rope and vertical was \(14.2^\circ\), we used the sine function to determine how the angle affects the vertical motion:\[h = L \times \sin(\theta)\]where \(L\) is the length of the rope and \(\theta\) is the angle from the vertical. This calculation allows us to understand how high the pendulum rises and how kinetic and potential energy interchange during its swing. Thus, trigonometry provides us with a clearer picture of pendulum dynamics by breaking down complex motion into comprehensible components.

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Most popular questions from this chapter

Consider two identical oscillators, each with spring constant \(k\) and mass \(m\), in simple harmonic motion. One oscillator is started with initial conditions \(x_{0}\) and \(v_{j}\) the other starts with slightly different conditions, \(x_{0}+\delta x\) and \(v_{0}+\delta v_{1}\) a) Find the difference in the oscillators' positions, \(x_{1}(t)-x_{2}(t)\) for all t. b) This difference is bounded; that is, there exists a constant \(C\) independent of time, for which \(\left|x_{1}(t)-x_{2}(t)\right| \leq C\) holds for all \(t\). Find an expression for \(C\). What is the best bound, that is, the smallest value of \(C\) that works? (Note: An important characteristic of chaotic systems is exponential sensitivity to initial conditions; the difference in position of two such systems with slightly different initial conditions grows exponentially with time. You have just shown that an oscillator in simple harmonic motion is not a chaotic system.)

A grandfather clock uses a physical pendulum to keep time. The pendulum consists of a uniform thin rod of mass \(M\) and length \(L\) that is pivoted freely about one end. with a solid sphere of the same mass, \(M,\) and a radius of \(L / 2\) centered about the free end of the rod. a) Obtain an expression for the moment of inertia of the pendulum about its pivot point as a function of \(M\) and \(L\). b) Obtain an expression for the period of the pendulum for small oscillations.

An \(80.0-\mathrm{kg}\) bungee jumper is enjoying an afternoon of jumps. The jumper's first oscillation has an amplitude of \(10.0 \mathrm{~m}\) and a period of \(5.00 \mathrm{~s}\). Treating the bungee cord as spring with no damping, calculate each of the following: a) the spring constant of the bungee cord. b) the bungee jumper's maximum speed during the ascillation, and c) the time for the amplitude to decrease to \(2.00 \mathrm{~m}\) (with air resistance providing the damping of the oscillations at \(7.50 \mathrm{~kg} / \mathrm{s})\)

The relative motion of two atoms in a molecule can be described as the motion of a single body of mass \(m\) moving in one dimension, with potcntial encrgy \(U(r)=A / r^{12}-B / r^{6}\) where \(r\) is the separation between the atoms and \(A\) and \(B\) are positive constants. a) Find the equilibrium separation, \(r_{0}\) of the atoms, in terms of the constants \(A\) and \(B\) b) If moved slightly, the atoms will oscillate about their equilibrium separation. Find the angular frequency of this oscillation, in terms of \(A, B,\) and \(m\).

If you kick a harmonic oscillator sharply, you impart to it an initial velocity but no initial displacement. For a weakly damped oscillator with mass \(m\), spring constant \(k\). and damping force \(F_{y}=-b v,\) find \(x(t),\) if the total impulse delivered by the kick is \(J_{0}\).

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