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What is the frequency of a simple pendulum \(2.0 \mathrm{~m}\) long (a) in a room, (b) in an elevator accelerating upward at a rate of \(2.0\) \(\mathrm{m} / \mathrm{s}^{2}\), and \((\mathrm{c})\) in free fall?

Short Answer

Expert verified
(a) \(0.353\,\mathrm{Hz};\) (b) \(0.388\,\mathrm{Hz};\) (c) \(0\,\mathrm{Hz}.\)

Step by step solution

01

Understanding the Pendulum's Frequency Formula

The frequency of a simple pendulum is given by the formula \( f = \, \frac{1}{2\pi} \sqrt{\frac{g}{L}} \), where \( g \) is the acceleration due to gravity and \( L \) is the length of the pendulum. In all scenarios, we'll use this formula to calculate frequency.
02

Calculate Frequency in a Room

In a room, the acceleration due to gravity \( g \) is approximately \( 9.81\, \mathrm{m/s}^2 \). Substituting the values into the formula gives: \[ f = \frac{1}{2\pi} \sqrt{\frac{9.81}{2.0}} \] Calculating this, we find \( f \approx 0.353\, \mathrm{Hz} \).
03

Calculate Frequency in an Elevator Accelerating Upward

In the elevator, the effective acceleration \( g' \) is the sum of gravitational acceleration and the elevator's acceleration, \( g' = 9.81 + 2.0 = 11.81\, \mathrm{m/s}^2 \). Substituting into the formula gives: \[ f = \frac{1}{2\pi} \sqrt{\frac{11.81}{2.0}} \] This results in \( f \approx 0.388\, \mathrm{Hz} \).
04

Calculate Frequency in Free Fall

In free fall, the effective gravitational acceleration \( g'' \) is 0 because both the elevator and the pendulum are accelerating downwards at the same rate. Thus, \( f = \frac{1}{2\pi} \sqrt{\frac{0}{2.0}} = 0\, \mathrm{Hz} \), meaning the pendulum doesn't oscillate.

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Key Concepts

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

Pendulum Length
The length of a pendulum, denoted as \( L \), is a critical factor in determining its frequency of oscillation. A simple pendulum consists of a mass (called the bob) that swings back and forth under the influence of gravity. The length, \( L \), specifically refers to the length of the string or rod from the pivot to the center of mass of the bob.For students studying pendulums, it’s important to understand that the length directly influences how quickly or slowly the pendulum swings. The pendulum length affects the period (\( T \)) of a pendulum’s swing, and is given by the formula:\[T = 2\pi \sqrt{\frac{L}{g}}\]This shows that the pendulum’s period is proportional to the square root of its length. As the length increases, the period becomes longer, leading to a lower frequency of oscillation. This relationship highlights one of the elegant principles of physics: how a simple change in an object’s dimensions can dramatically affect its behavior.
Gravitational Acceleration
Gravitational acceleration, often denoted by \( g \), is a measure of how fast an object speeds up as it falls freely under the influence of gravity. On Earth, this value is approximately \( 9.81 \, \mathrm{m/s}^2 \), but it can vary slightly depending on geographic location and altitude.In the context of a pendulum, gravitational acceleration is fundamental in determining the pendulum’s frequency. The formula for the frequency \( f \) of a simple pendulum contains \( g \):\[f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}\]This shows how gravitational acceleration affects the pendulum’s oscillation. If \( g \) increases, the frequency increases, meaning the pendulum swings faster. Conversely, a smaller \( g \) results in a slower oscillation. Understanding how gravitational acceleration interacts with other factors is key to mastering concepts related to pendulums.
Effective Gravitational Acceleration
In physics problems involving oscillations, effective gravitational acceleration becomes an essential concept, especially when external forces are involved. It is represented as \( g' \) and expresses the total acceleration acting on a system, combining standard gravitational acceleration with additional forces.Consider a pendulum inside an elevator accelerating upward. The effective gravitational acceleration \( g' \) increases because the elevator’s upward acceleration adds to Earth’s gravity:\[g' = g + a\]Thus, if an elevator accelerates upward at \( 2.0 \, \mathrm{m/s}^2 \), \( g' \) becomes \( 11.81 \, \mathrm{m/s}^2 \). This affects how rapidly the pendulum swings and increases its frequency of oscillation:\[f = \frac{1}{2\pi} \sqrt{\frac{11.81}{2.0}}\]In free fall, however, both the pendulum and the elevator fall with the same acceleration, making \( g' = 0 \). This renders the pendulum stationary in terms of oscillation frequency, illustrating how drastically changes in effective gravitational acceleration can impact pendulum motion.
Oscillation
Oscillation refers to the repetitive back-and-forth motion of an object, like a pendulum swinging. It’s a key concept not only in physics but in numerous fields such as engineering and music.A simple pendulum exhibits harmonic oscillation, meaning its motion can be predicted based on simple mathematical relationships. Key elements defining oscillation include the amplitude, period, and frequency:- **Amplitude** is the maximum extent of the swing from its equilibrium position.- **Period \( T \)** refers to the time it takes to complete one full cycle of motion.- **Frequency \( f \)** is the number of cycles per second.Understanding oscillation involves recognizing how changes in physical conditions affect these properties. External factors like effective gravitational acceleration can alter the frequency, demonstrating the responsiveness of oscillating systems to environmental changes. The elegant movement of the pendulum beautifully exhibits the periodic nature of oscillations, making these principles tangible and enlightening for students.

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Most popular questions from this chapter

An oscillating block-spring system takes \(0.75 \mathrm{~s}\) to begin repeating its motion. Find (a) the period, (b) the frequency in hertz, and (c) the angular frequency in radians per second.

An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The springs of a certain car are adjusted so that the oscillations have a frequency of \(3.00 \mathrm{~Hz}\). (a) What is the spring constant of each spring if the mass of the car is \(1450 \mathrm{~kg}\) and the mass is evenly distributed over the springs? (b) What will be the oscillation frequency if five passengers, averaging \(73.0 \mathrm{~kg}\) each, ride in the car with an even distribution of mass?

A loudspeaker produces a musical sound by means of the oscillation of a diaphragm whose amplitude is limited to \(1.00 \mu \mathrm{m}\). (a) At what frequency is the magnitude \(a\) of the diaphragm's acceleration equal to \(g\) ? (b) For greater frequencies, is \(a\) greater than or less than \(g\) ?

An oscillating block-spring system has a mechanical energy of \(1.00 \mathrm{~J}\), an amplitude of \(10.0 \mathrm{~cm}\), and a maximum speed of \(1.20\) \(\mathrm{m} / \mathrm{s}\). Find (a) the spring constant, (b) the mass of the block, and (c) the frequency of oscillation.

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. 15-38). 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.

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