Chapter 7: Problem 99
A tightrope walker of mass \(m\) stands at rest midway along a cable of length \(L\) and negligible mass. The cable is stretched tightly between two supports, giving it a tension \(F\). If this equilibrium is disturbed, the tightrope walker undergoes smallamplitude vertical oscillations. Show that the period of these oscillations is \(T=2 \pi \sqrt{\frac{m L}{4 F}}\).
Short Answer
Step by step solution
Understand the scenario
Use the small-angle approximation
Establish the forces and dynamics
Derive the period formula
Relate tension to the spring constant
Substitute the spring constant into the period formula
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Tightrope Walker
Period of Oscillation
- It uses the known relationship from simple harmonic motion: \[ T = 2 \pi \sqrt{\frac{m}{k}} \]where \( k \) is the effective spring constant.
- The specific period for our problem is: \[ T = 2\pi \sqrt{\frac{mL}{4F}} \]This means the period increases with mass and length but decreases with greater tension in the cable.
Small-Angle Approximation
- It allows us to model the vertical oscillations as simple harmonic motion easier.
- The approximation brings simplification to trigonometric terms in the equations of motion that describe the walker’s path.
Tension in Cable
- The tension acts horizontally, and when combined with gravity, it creates a vertical component that brings the walker back to balance.
- When viewed as a form of spring, the cable's tension directly affects the oscillation period by acting as the effective spring constant \( k = \frac{4F}{L} \).