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For a simple pendulum, find the angular amplitude \(\theta_{m}\) at which the restoring torque required for simple harmonic motion deviates from the actual restoring torque by \(1.0 \%\). (See "Trigonometric Expansions" in Appendix E.)

Short Answer

Expert verified
The angular amplitude is approximately 0.346 radians.

Step by step solution

01

Understand Torque in a Pendulum

For a simple pendulum, the actual restoring torque is given by \( \tau = -mgL\sin(\theta) \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, \( L \) is the length of the pendulum, and \( \theta \) is the angular displacement. For simple harmonic motion (SHM), this is approximated by \( \tau_{SHM} = -mgL\theta \).
02

Calculate First-Order Approximation

By using the Taylor series expansion for small angles, \( \sin(\theta) \approx \theta - \frac{\theta^3}{6} \). For small \( \theta \), the first order approximation is \( \sin(\theta) \approx \theta \). But, this approximation begins to deviate as \( \theta \) increases.
03

Determine Percentage Deviation

We want the deviation to be 1%. Thus, the condition is \( |\ \tau_{actual} - \tau_{SHM} \ | / \tau_{actual} = 0.01 \). Simplifying gives \( | mgL\theta - mgL\theta + mgL\frac{\theta^3}{6} | / mgL\sin(\theta) = 0.01 \).
04

Solve for \( \theta_m \)

Substitute the approximations for \( \sin(\theta) \) using the Taylor expansion: \[ \left| \frac{mgL\frac{\theta^3}{6}}{mgL\left(\theta - \frac{\theta^3}{6}\right)} \right| = 0.01 \]. Simplifying this equation, cancel terms, and solve for \( \theta \). This results in \( \theta^2 = 0.12 \times \theta \), resulting in \( \theta_m \approx 0.346 \) radians.

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Key Concepts

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

Angular Amplitude
Angular amplitude refers to the maximum angular displacement of a pendulum from the vertical rest position. Essentially, it is how far the pendulum swings away from its central, or equilibrium, position. Angular amplitude is crucial because it determines the energy stored in the pendulum's motion. The larger the amplitude, the more energy. In simple harmonic motion, which we will discuss shortly, energy conservation and small angle approximations are often used to simplify calculations.
  • For a simple pendulum, the amplitude is usually small enough such that the sine of the angle can be approximated as the angle itself (&\( \sin(\theta) \approx \theta \) for small \( \theta \)).
  • However, as angles increase, this approximation becomes less accurate, leading to differences between actual motion and the idealized simple harmonic motion.
Understanding angular amplitude is key when determining how these deviations come into play.
Restoring Torque
In a simple pendulum, restoring torque is what brings the pendulum back to its equilibrium position. When a pendulum swings away from the vertical, gravity creates this torque, aiming to pull it back to the center. The restoring torque for a pendulum is calculated by (&\( \tau = -mgL\sin(\theta) \)), where \( m \) is the mass, \( g \) is the gravitational acceleration, \( L \) is the length of the pendulum, and \( \theta \) is the angle of displacement. The minus sign indicates that the torque acts in the direction opposite to the displacement.
  • For small angles, it's approximated by (&\( \tau_{SHM} = -mgL\theta \)), assuming simple harmonic motion.
  • This approximation works well when \( \theta \) is small because \( \sin(\theta) \approx \theta \).
  • The deviation from simple harmonic motion starts becoming significant as \( \theta \) increases, which would require adjusting calculations for accuracy.
Restoring torque plays a central role in pendulum dynamics by dictating how the pendulum returns to its rest position.
Simple Harmonic Motion
Simple harmonic motion, or SHM, is a type of periodic oscillatory motion where the force acting to restore an object to its equilibrium position is directly proportional to the displacement and acts in the opposite direction. It provides a simplified model of oscillations where, particularly in pendulums, can describe the behavior under assumptions of small angles.
  • For a pendulum in SHM, the restoring torque equation simplifies from (&\( \tau = -mgL\sin(\theta) \)) to (&\( \tau_{SHM} = -mgL\theta \)) under small angle approximations.
  • SHM assumes that the motion is a perfect sine wave, signifying constant energy distribution across its cycle.
  • Any deviation in the actual restoring force from this approximation leads to the need to consider modifications, especially for angles where such simplifications break down.
Using these SHO principles helps in efficiently predicting pendulum motions with minimal computational complexity.
Taylor Series Expansion
The Taylor Series Expansion is a mathematical technique used to approximate more complex functions through a series of simpler polynomial terms. In pendulum physics, we use it to simplify the equation for restoring torque when dealing with small angles. For example, the sine function is replaced by its Taylor expansion: (&\( \sin(\theta) \approx \theta - \frac{\theta^3}{6} \)) for small \( \theta \). This approximation significantly aids in the analysis by reducing the problem's complexity when using \( \theta \) alone instead of \( \sin(\theta) \).
  • The first-order approximation (&\( \sin(\theta) \approx \theta \)) is used in conditions where high precision isn't necessary and \( \theta \) is minimal.
  • In cases requiring more accuracy, such as the calculation of angular amplitude at which torque deviation reached 1%, the additional term \( \frac{\theta^3}{6} \) helps capture the deviation from true simple harmonic motion.
  • This method is crucial in simplifying equations to make practical theoretical predictions about a pendulum's behavior.
The Taylor Series provides the toolkit necessary for progressing from theory to real-world application in pendulum analysis.

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Most popular questions from this chapter

A block of mass \(M=5.4\) \(\mathrm{kg}\), at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant \(k=6000 \mathrm{~N} / \mathrm{m}\). A bullet of mass \(m=9.5 \mathrm{~g}\) and velocity \(\vec{v}\) of magnitude \(630 \mathrm{~m} / \mathrm{s}\) strikes and is embedded in the block (Fig. 15. 40). Assuming the compression of the spring is negligible until the bullet is embedded, determine (a) the speed of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.

A block weighing \(20 \mathrm{~N}\) oscillates at one end of a vertical spring for which \(k=100 \mathrm{~N} / \mathrm{m} ;\) the other end of the spring is attached to a ceiling. At a certain instant the spring is stretched \(0.30 \mathrm{~m}\) beyond its relaxed length (the length when no object is attached) and the block has zero velocity. (a) What is the net force on the block at this instant? What are the (b) amplitude and (c) period of the resulting simple harmonic motion? (d) What is the maximum kinetic energy of the block as it oscillates?

A block weighing \(10.0 \mathrm{~N}\) is attached to the lower end of a vertical spring \((k=200.0 \mathrm{~N} / \mathrm{m})\), the other end of which is attached to a ceiling. The block oscillates vertically and has a kinetic energy of \(2.00 \mathrm{~J}\) as it passes through the point at which the spring is unstretched. (a) What is the period of the oscillation? (b) Use the law of conservation of energy to determine the maximum distance the block moves both above and below the point at which the spring is unstretched. (These are not necessarily the same.) (c) What is the amplitude of the oscillation? (d) What is the maximum kinetic energy of the block as it oscillates?

The scale of a spring balance that reads from 0 to \(15.0 \mathrm{~kg}\) is \(12.0 \mathrm{~cm}\) long. A package suspended from the balance is found to oscillate vertically with a frequency of \(2.00 \mathrm{~Hz}\). (a) What is the spring constant? (b) How much does the package weigh?

Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. What is their phase difference?

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