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Chapter 13: Problem 43

A building in San Francisco has light fixtures consisting of small \(2.35-\mathrm{kg}\) bulbs with shades hanging from the ceiling at the end of light thin cords \(1.50 \mathrm{~m}\) long. If a minor earthquake occurs, how many swings per second will these fixtures make?

Short Answer

Expert verified
The fixtures make approximately 0.407 swings per second.

Step by step solution

01

Identify Given Values and Known Formula

We are given the mass of the bulb as \(2.35\, \text{kg}\), the length of the cord as \(1.50\, \text{m}\), and are asked to find the number of swings per second (frequency) of the pendulum. We will use the formula for the period of a simple pendulum: \(T = 2\pi\sqrt{\frac{L}{g}}\), where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity.
02

Calculate the Period of the Pendulum

First, we assume standard gravity \(g = 9.81\, \text{m/s}^2\). Using the period formula \(T = 2\pi\sqrt{\frac{L}{g}}\), substitute the given values, \(L = 1.50\, \text{m}\): \[T = 2\pi\sqrt{\frac{1.50}{9.81}}.\]
03

Solve for the Period

Calculate \(T\): \[T = 2\pi\sqrt{\frac{1.50}{9.81}} \approx 2\pi\sqrt{0.153} \approx 2\pi \times 0.3913 \approx 2.46\, \text{seconds}.\]
04

Calculate the Frequency of the Pendulum

Frequency \(f\) is the reciprocal of the period. Use the formula \(f = \frac{1}{T}\): \[f = \frac{1}{2.46} \approx 0.407 \text{Hz}.\]
05

Interpret the Result

The frequency of \(0.407 \text{Hz}\) means that the pendulum (light fixture) makes approximately 0.407 swings per second during the minor earthquake.

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Key Concepts

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

Period of a Pendulum
The period of a pendulum is the time it takes for the pendulum to complete one full swing back and forth. This concept is crucial in understanding pendulum motion, as it directly relates to how often the pendulum oscillates. For a simple pendulum, which consists of a weight (or bob) attached to a string or rod of negligible mass, the period can be calculated using:\[ T = 2\pi \sqrt{\frac{L}{g}} \]Where:
  • \( T \) is the period of the pendulum in seconds.
  • \( L \) is the length of the pendulum in meters.
  • \( g \) is the acceleration due to gravity in meters per second squared (\(9.81 \, \text{m/s}^2\) on Earth).
The formula highlights that the period depends solely on the pendulum length and the local gravity, not the mass of the bob. This means that as the length increases, so does the period, causing the pendulum to swing more slowly.
Acceleration Due to Gravity
The acceleration due to gravity, denoted as \( g \), plays a pivotal role in pendulum motion. On Earth, \( g \) is approximately \( 9.81 \, \text{m/s}^2 \). This value represents how quickly an object accelerates downwards due to gravity.In the context of a pendulum:
  • The acceleration due to gravity is responsible for pulling the pendulum downwards.
  • It ensures that pendulums return to their resting position and begin their oscillatory motion.
  • It is a fundamental constant used in determining the pendulum's period.
While \( g \) remains fairly constant, its influence is noticeable in different gravitational environments (like on other planets), which would alter the pendulum's period accordingly.
Pendulum Motion
Pendulum motion describes the repetitive back-and-forth movement of a pendulum. This motion is governed by restoring forces that act to return the pendulum to its equilibrium position.Key elements of pendulum motion include:
  • Simple Harmonic Motion: When the angles of deflection are small, the pendulum exhibits simple harmonic motion, where the restoring force is proportional to its displacement.
  • Amplitude and Energy: The amplitude is the maximum extent of the swing, while energy in a pendulum system is conserved, transitioning between potential and kinetic energy.
  • Period and Frequency: As derived earlier, the period \( T \) is the time for one cycle and the frequency \( f \) (\( f = \frac{1}{T} \)) is how often the pendulum swings per second.
Understanding pendulum motion aids in comprehending various phenomena in physics, engineering, and even in natural settings, such as how clocks work or buildings sway during earthquakes.
Earthquake Effects on Pendulums
During an earthquake, pendulums can be used to measure the intensity of seismic activity. As the ground shakes, so does the fixture holding pendulums, causing them to swing. Effects during an earthquake:
  • Sensitivity: Pendulums are sensitive to motion, making them useful in detecting and measuring seismic activity.
  • Frequency Increase: As the building sways, the swinging pendulum can provide data on how quickly it moves, termed as its frequency.
  • Resonance: The natural frequency of a pendulum might match the frequency of the building's vibrations, leading to resonance and larger swings.
These dynamics help engineers and scientists to design more earthquake-resilient structures and to assess a building's response to seismic events. Understanding how pendulums react to earthquakes can improve safety and energy absorption mechanisms in buildings.

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Most popular questions from this chapter

A spring of negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) is hung vertically, and a \(0.200-\mathrm{kg}\) pan is suspended from its lower end. A butcher drops a 2.2\(\cdot \mathrm{kg}\) steak onto the pan from a height of 0.40 \(\mathrm{m}\) . The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What are (a) the speed of the pan and steak imto immediately after the collision; (b) the amplitude of the subsequent motion; (c) the period of that motion?

Show that the expression for the period of a physical penduhm reduces to that of a simple pendulum if the physical pendulum consists of a particle with mass \(m\) on the end of a massless string of length \(L .\)

A \(0.400-\mathrm{kg}\) object undergoing \(\mathrm{SHM}\) has \(a_{x}=-2.70 \mathrm{m} / \mathrm{s}^{2}\) when \(x=0.300 \mathrm{m}\) . What is the time for one oscillation?

Weighing Astronauts. This procedure has actually been used to "weigh" astronauts in space. A \(42.5-\mathrm{kg}\) chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes 1.30 s to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes 2.54 s for one cycle. What is the mass of the astronaut?

(a) What is the change \(\Delta T\) in the period of a simple penduIum when the acceleration of gravity \(g\) changes by \(\Delta g ?\) (Hint: The new period \(T+\Delta T\) is obtained by substituting \(g+\Delta g\) for \(g :\) $$\boldsymbol{T}+\Delta \boldsymbol{T}=2 \pi \sqrt{\frac{\boldsymbol{L}}{\boldsymbol{g}+\Delta \boldsymbol{g}}}$$ To obtain an approximate expression, expand the factor \((g+\Delta g)^{-1 / 2}\) using the binomial theorem (Appendix B) and keep only the first two terms: $$(g+\Delta g)^{-1 / 2}=g^{-1 / 2}-\frac{1}{2} g^{-3 / 2} \Delta g+\cdots$$ The other terms contain higher powers of \(\Delta g\) and are very small if \(\Delta g\) is small.) Express your result as the fractional change in period \(\Delta T / T\) in terms of the fractional change \(\Delta g / g .\) (b) A pendulum clock keeps correct time at a point where \(g=9.8000 \mathrm{m} / \mathrm{s}^{2},\) but is found to lose 4.0 \(\mathrm{s}\) each day at a higher elevation. Use the result of part (a) to find the approximate value of \(g\) at this new location.
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