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Chapter 10: Problem 83

Multiple-Concept Example 6 reviews the principles that play roles in this problem. A bungee jumper, whose mass is 82 kg, jumps from a tall platform. After reaching his lowest point, he continues to oscillate up and down, reaching the low point two more times in 9.6 s. Ignoring air resistance and assuming that the bungee cord is an ideal spring, determine its spring constant.

Short Answer

Expert verified
The spring constant is approximately 242 N/m.

Step by step solution

01

Identify the Concept

This problem involves energy conservation and oscillatory motion. Specifically, we're dealing with a situation where the bungee cord behaves like an ideal spring, making this a simple harmonic motion problem.
02

Determine the Period of Oscillation

We know that the jumper reaches the lowest point twice after the initial jump in a period of 9.6 s. This means the oscillation period is \[ T = \frac{9.6 \text{ s}}{2} = 4.8 \text{ s} \] since each cycle involves one full oscillation from the lowest point to the highest point and back again.
03

Use the Formula for the Period

For a mass-spring system, the period \( T \) is given by \[ T = 2\pi \sqrt{\frac{m}{k}} \]where \( m \) is the mass and \( k \) is the spring constant. Substitute the known values into the equation:\[ 4.8 = 2\pi \sqrt{\frac{82}{k}} \]
04

Solve for the Spring Constant \( k \)

Rearrange the equation from Step 3 to solve for \( k \):\[ \sqrt{\frac{82}{k}} = \frac{4.8}{2\pi} \]\[ \frac{82}{k} = \left(\frac{4.8}{2\pi}\right)^2 \]\[ k = \frac{82}{\left(\frac{4.8}{2\pi}\right)^2} \]Calculate:\[ k \approx \frac{82}{0.5820^2}\]\[ k \approx \frac{82}{0.3387}\]\[ k \approx 242 \, \text{N/m} \]
05

Conclusion

Thus, the spring constant of the bungee cord is approximately 242 N/m. This represents the stiffness of the bungee cord in restoring force as a spring.

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Key Concepts

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

Energy Conservation
When we talk about energy conservation in simple harmonic motion, it's about how energy transforms between different forms. In a mass-spring system like the bungee jumper, energy shifts between potential and kinetic forms without any loss because there's no air resistance.
  • At the highest points of the jump, the potential energy is at its maximum since the jumper is at a maximum distance from the equilibrium.
  • As the jumper falls, potential energy is converted into kinetic energy.
  • At the lowest point, kinetic energy is maximized while potential energy in the spring (elastic potential energy) is also significant here.
In ideal conditions, the total mechanical energy stays constant. By using equations related to energy conservation, you can understand the motion better and even predict the jumper's speed at various points. But remember, this ideal behavior is simplified, as real-life scenarios generally include some energy loss.
Mass-Spring System
In physics, a mass-spring system is a classic example of simple harmonic motion. It's where a mass is attached to a spring and allowed to move back and forth about an equilibrium position. The bungee jumper scenario models this concept very well.
  • "Mass" represents the jumper here, weighing 82 kg in this problem.
  • "Spring" refers to the bungee cord, which behaves like an ideal spring with elasticity.
  • The spring constant "k" describes the stiffness of the bungee cord. In our case, it was calculated to be approximately 242 N/m.
The oscillation you see as the jumper moves up and down is governed by the properties of this mass-spring system. Despite the simplicity of this model, it captures key insights into the motion involved in bungee jumping.
Period of Oscillation
The period of oscillation is a critical feature in understanding simple harmonic motion. It's the time taken for one complete cycle of movement, from one extreme to the other, and back again. For the bungee jumper:
  • The time given for reaching the lowest point twice was 9.6 seconds.
  • Since it happens two times in this span, the single oscillation period reflects the round trip from high to low and back, which is calculated as 4.8 seconds.
The formula to determine the period in a mass-spring system is \[ T = 2\pi \sqrt{\frac{m}{k}} \] This period calculation reflects how mass and spring constant interrelate, as changes in these will alter the time duration of one complete oscillation. Understanding this helps you predict how different jumper weights or cord stiffnesses affect the motion.
Bungee Jumping Physics
Bungee jumping involves a fascinating interplay of physics principles, mainly modeled as a mass-spring system in this context. Let's explore:
  • **Elastic Potential Energy**: The bungee cord, acting like a spring, stores energy when stretched.
  • **Gravitational Potential Energy**: It decreases as the jumper falls and increases again during ascent.
  • **Kinetic Energy**: Peaks mid-journey; it's highest at the midpoint of travel.
Bungee jumping showcases simple harmonic motion's dynamics. In a real jump, factors like air resistance cause energy losses, leading to dampened oscillations. But the idealized form allows us to cement foundational principles that are transferable across many physics applications, making bungee jumping an exciting example for educators and thrill-seekers alike.

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Most popular questions from this chapter

An 86.0-kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of \(1.20 \times 10^{3} \mathrm{N} / \mathrm{m} .\) He accidentally slips and falls freely for \(0.750 \mathrm{m}\) before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

A uniform \(1.4-\mathrm{kg}\) rod that is \(0.75 \mathrm{m}\) long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are \(59 \mathrm{N} / \mathrm{m}\) and \(33 \mathrm{N} / \mathrm{m}\). Find the angle that the rod makes with the horizontal.

A spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is \(0.80 \mathrm{kg}\), and the spring has a spring constant of \(59 \mathrm{N} / \mathrm{m}\). The coefficient of static friction between the box and the table on which it rests is \(\mu_{\mathrm{s}}=0.74 .\) How far can the spring be stretched from its unstrained position without the box moving when it is released?

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 to \(3.0 \times 10^{-2} \mathrm{N}\). The length and radius of the collagen are, respectively, 2.5 and \(0.091 \mathrm{cm},\) and Young's modulus is \(3.1 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}\). (a) If the stretching obeys Hooke's law, what is the spring constant \(k\) for collagen? (b) How much work is done by the variable force that stretches the collagen? (See Section 6.9 for a discussion of the work done by a variable force.)

A 1.1 -kg object is suspended from a vertical spring whose spring constant is \(120 \mathrm{N} / \mathrm{m}\). (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of \(0.20 \mathrm{m}\) and released from rest. Find the speed with which the object passes through its original position on the way up.
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