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Chapter 16: Problem 44

44\. In a Car A simple pendulum of length \(L\) and mass \(m\) is suspended in a car that is traveling with constant speed \(|\vec{v}|\) around a circle of radius \(R\). If the pendulum undergoes small oscillations in a radial direction about its equilibrium position, what will be its frequency of oscillation?

Short Answer

Expert verified
The frequency of oscillation is given by \(f = \frac{1}{2\pi} \sqrt{\frac{g}{L} + \frac{v^2}{R L}}\).

Step by step solution

01

- Identify the forces on the pendulum

The pendulum experiences a gravitational force downward, given by \(\vec{F_g} = m\vec{g}\), and a centripetal force \(\vec{F_c} = \frac{mv^2}{R}\) directed toward the center of the circular path.
02

- Determine the equilibrium position

In the car's rotating frame, the pendulum will hang at an angle \(\theta\) such that the sum of the forces in the radial direction equals zero. This means that \(T \sin(\theta) = \frac{mv^2}{R}\) and \(T \cos(\theta) = mg\), where \(T\) is the tension in the string.
03

- Relate the tensions and angles

Dividing \(\frac{T \sin(\theta)}{T \cos(\theta)} = \frac{\frac{mv^2}{R}}{mg} \) simplifies to \(\tan(\theta) = \frac{v^2}{Rg}\). For small oscillations, \(\sin(\theta) \approx \theta\) and \(\cos(\theta) \approx 1\), so \(\theta = \frac{v^2}{Rg}\).
04

- Set up the equation for radial oscillations

The restoring force for small radial displacements \(\delta L\) from equilibrium is given by \(F = -m \left( \frac{g}{L} + \frac{v^2}{R L} \right) \delta L\). This can be written as a simple harmonic motion equation, \(F = -k \delta L\), where \(k = m \left( \frac{g}{L} + \frac{v^2}{R L} \right)\).
05

- Find the angular frequency \(\omega\)

The angular frequency \(\omega\) for simple harmonic motion is given by \(\omega = \sqrt{\frac{k}{m}}\). Substituting the value of \(k\) from the previous step, we get \(\omega = \sqrt{\frac{g}{L} + \frac{v^2}{R L}}\).
06

- Calculate the frequency of oscillation

The frequency \(f\) is related to the angular frequency \(\omega\) by \(f = \frac{\omega}{2\pi}\). Thus, \(f = \frac{1}{2\pi} \sqrt{\frac{g}{L} + \frac{v^2}{R L}}\).

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Key Concepts

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

Radial Oscillations
When we talk about radial oscillations, we refer to tiny back-and-forth movements along a line that extends from the center of a circle outward. In our pendulum scenario, these oscillations occur due to slight deviations from the pendulum's equilibrium position inside the moving car. Imagine the pendulum swinging toward and away from the center of the circular path鈥攖he direction along this swing is radial. Understanding radial oscillations is crucial because it gives us a foundation for analyzing the forces and movements of pendulums and similar systems in rotational contexts.
Centripetal Force
Centripetal force is a force that always points toward the center of a circle, keeping an object moving along a circular path. For our pendulum in the car, the centripetal force is the tension in the string that directs part of its movement toward the circle's center. The formula for centripetal force is \(\frac{mv^2}{R}\), where \(m\) is the mass of the pendulum, \(v\) is the car's speed, and \(R\) is the circle's radius. Without this force, the pendulum would simply move in a straight line due to inertia. Knowing this force helps us understand why and how the pendulum shifts as the car navigates the circular track.
Simple Harmonic Motion
Simple harmonic motion (SHM) describes a type of periodic movement, like our pendulum swinging back and forth around an equilibrium position. In our case, the pendulum demonstrates SHM due to the restoring forces acting on it when displaced. The force formula looks like this: \(F = -k \delta L\), with \(\delta L\) as the displacement from equilibrium and \(k\) the effective 'spring constant' for the pendulum. This setup tells us that the pendulum will oscillate in predictable cycles, making it easier to calculate the frequency and understand its overall behavior within the moving car.
Equilibrium Position
The equilibrium position of the pendulum is the spot where all forces balance out, resulting in no net force acting on it. For the pendulum in our exercise, this position forms an angle \(\theta\) with the vertical axis due to the gravitational force pulling it down and the centripetal force pulling it toward the center of the circular path. By understanding that \(\tan(\theta) = \frac{v^2}{Rg}\), we can find this equilibrium position accurately. Knowing this spot is essential because it serves as the central point from which we measure oscillations and predict how the pendulum will behave.
Angular Frequency
Angular frequency \(\omega\) tells us how quickly the pendulum swings during its oscillations. For simple harmonic motion, \(\omega\) links to the restoring force and can be derived using the formula \(\omega = \sqrt{\frac{k}{m}}\). In our pendulum scenario, we find \(\omega = \sqrt{\frac{g}{L} + \frac{v^2}{R L}}\). This value helps us directly calculate the oscillation frequency using \(f = \frac{\omega}{2\pi}\). Understanding angular frequency is vital because it impacts how fast the pendulum oscillates, which influences the time it takes for each swing and our overall analysis of the system's dynamic behavior.

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Most popular questions from this chapter

56\. Vertical Spring A vertical spring stretches \(9.6 \mathrm{~cm}\) when a \(1.3 \mathrm{~kg}\) block is hung from its end. (a) Calculate the spring constant. This block is then displaced an additional \(5.0 \mathrm{~cm}\) downward and released from rest. Find (b) the period, (c) the frequency, (d) the amplitude, and (e) the maximum speed of the resulting SHM.

22\. Two Particles 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?

40\. Pendulum with Disk A uniform circular disk whose radius \(R\) is \(12.5 \mathrm{~cm}\) is suspended as a physical pendulum from a point on its rim. (a) What is its period? (b) At what radial distance \(r

46\. Angular Amplitude For a simple pendulum, find the angular amplitude \(\Theta\) 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.)

58\. Lightly Damped The amplitude of a lightly damped oscillator decreases by \(3.0 \%\) during each cycle. What fraction of the mechanical energy of the oscillator is lost in each full oscillation?
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