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?

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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