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The length of a simple pendulum is \(0.79 \mathrm{m}\) and the mass of the particle (the "bob") at the end of the cable is \(0.24 \mathrm{kg}\). The pendulum is pulled away from its equilibrium position by an angle of \(8.50^{\circ}\) and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing?

Short Answer

Expert verified
(a) 3.52 rad/s, (b) 0.021 J, (c) 0.42 m/s.

Step by step solution

01

Calculate Angular Frequency

To find the angular frequency of the simple pendulum, we use the formula for angular frequency for simple harmonic motion: \[ \omega = \sqrt{\frac{g}{L}} \]where \( g \approx 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( L = 0.79 \, \text{m} \) is the length of the pendulum. Substitute these values into the formula:\[ \omega = \sqrt{\frac{9.81}{0.79}} = \sqrt{12.42} = 3.52 \, \text{rad/s} \].
02

Calculate the Total Mechanical Energy

The total mechanical energy of a pendulum in simple harmonic motion consists only of potential and kinetic energy. At its maximum displacement (amplitude), all energy is potential:\[ E = mgh \]Calculate the height \( h \) the bob reaches using the initial angle \( \theta = 8.50^{\circ} \). The height can be found by\[ h = L - L\cos(\theta) = 0.79 - 0.79\cos(8.50^{\circ}) \approx 0.79 - 0.79 \times 0.989 = 0.0089 \, \text{m} \].Thus, substitute \( m = 0.24 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( h = 0.0089 \, \text{m} \) into the potential energy formula:\[ E = 0.24 \times 9.81 \times 0.0089 \approx 0.021 \, \text{J} \].
03

Calculate the Speed at the Lowest Point

At the lowest point of its swing, all the potential energy of the pendulum converts to kinetic energy:\[ \frac{1}{2}mv^2 = E \]Solve for the speed \( v \) by substituting the known values:\[ \frac{1}{2} \times 0.24 \times v^2 = 0.021 \rightarrow v^2 = \frac{0.021 \times 2}{0.24} = \frac{0.042}{0.24} = 0.175 \rightarrow v = \sqrt{0.175} \approx 0.42 \, \text{m/s} \].

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Key Concepts

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

Pendulum
Imagine swinging back and forth in a gentle rhythm; that's what a pendulum does. A pendulum is a simple device consisting of a weight, or bob, attached to the end of a string or rod that swings around a pivot point. Think of a grandfather clock or a child's swing. These motions are an excellent example of simple harmonic motion. In our exercise, the pendulum has a length of 0.79 meters and a bob with a mass of 0.24 kilograms. Even a slight angle, like 8.50 degrees, can initiate its smooth swing.

During this oscillation, the pendulum stores and transfers energy back and forth between potential and kinetic forms. As it moves away from its central position, the pendulum gains height, increasing potential energy. When it returns, swinging through the lowest point, potential energy turns into kinetic energy. This constant energy transfer is the magic behind the pendulum's perpetual, frictionless dance.
Angular Frequency
In the context of simple harmonic motion, angular frequency is a vital concept. It tells us how quickly the pendulum swings back and forth. Angular frequency is represented by the symbol \( \omega \) and is measured in radians per second.

For our pendulum, we use the formula: \[ \omega = \sqrt{\frac{g}{L}} \] where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity and \( L \) is the length of the pendulum. Substituting our given values, \( \omega = \sqrt{\frac{9.81}{0.79}} \approx 3.52 \, \text{rad/s} \).

This frequency indicates that every second, the pendulum completes a portion of its swing, precisely 3.52 radians worth. By understanding angular frequency, we set the pace for the swinging pendulum, allowing us to calculate other essential properties, like energy.
Mechanical Energy
Mechanical energy in a pendulum is the sum of its potential and kinetic energy. It remains constant if we neglect friction. Initially, when the pendulum is pulled aside, it gains potential energy due to its height.

To find the total mechanical energy, we calculate this potential energy at maximum displacement: \[ E = mgh \] Here, \( h \) is the height calculated from its vertical rise: \[ h = L - L\cos(\theta) \approx 0.0089 \, \text{m} \] Thus, the energy is: \[ E = 0.24 \times 9.81 \times 0.0089 \approx 0.021 \, \text{J} \].

This mechanical energy primarily exists as potential when highest and entirely converts to kinetic at the lowest swing point. This constant energy transformation is a core characteristic of the pendulum's motion.
Kinetic Energy
Kinetic energy is what the pendulum has as it swings fastest through its lowest point. It's the energy of motion, calculated with the formula: \[ \frac{1}{2}mv^2 = E \] At the lowest point, all the potential energy converts into kinetic energy. To find the speed of the bob, \( v \), we use the total mechanical energy: \[ v^2 = \frac{0.042}{0.24} = 0.175 \quad \Rightarrow \quad v = \sqrt{0.175} \approx 0.42 \, \text{m/s} \].

This calculation shows how fast the bob moves as it passes through the lowest point.

Understanding the relationship between kinetic energy and the pendulum's speed helps us appreciate how energy is conserved throughout its motion. It's a beautiful demonstration of nature's efficiency and the predictability of physics in action.

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Most popular questions from this chapter

A loudspeaker diaphragm is producing a sound for 2.5 s by moving back and forth in simple harmonic motion. The angular frequency of the motion is \(7.54 \times 10^{4} \mathrm{rad} / \mathrm{s} .\) How many times does the diaphragm move back and forth?

A 15.0-kg block rests on a horizontal table and is attached to one end of a mass less, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of 5.00 m/s in 0.500 s. In the process, the spring is stretched by 0.200 m. The block is then pulled at a constant speed of 5.00 m/s, during which time the spring is stretched by only 0.0500 m. Find (a) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.

A spring \((k=830 \mathrm{N} / \mathrm{m})\) is hanging from the ceiling of an elevator, and a 5.0-kg object is attached to the lower end. By how much does the spring stretch (relative to its unstrained length) when the elevator is accelerating upward at \(a=0.60 \mathrm{m} / \mathrm{s}^{2} ?\)

A 0.60-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 to 0.23 m (relative to its unstrained length), the speed of the sphere decreases from 5.70 to 4.80 m/s. What is the spring constant of the spring?

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is \(f=3.0 \mathrm{Hz} ?\)

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