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Hanging from a horizontal beam are nine simple pendulums of the following lengths: (a) \(0.10\), (b) \(0.30,(\mathrm{c}) 0.40,(\) d) \(0.80\), (e) \(1.2\), (f) \(2.8,(\mathrm{~g}) 3.5,(\mathrm{~h}) 5.0\), and (i) \(6.2 \mathrm{~m}\). Suppose the beam undergoes horizontal oscillations with angular frequencies in the range from \(2.00 \mathrm{rad} / \mathrm{s}\) to \(4.00 \mathrm{rad} / \mathrm{s}\). Which of the pendulums will be (strongly) set in motion?

Short Answer

Expert verified
Pendulums (d) and (e) will be strongly set in motion.

Step by step solution

01

Understanding the relationship between pendulum length and frequency

The frequency at which a pendulum naturally oscillates is determined by its length. For simple pendulums, the angular frequency \( \omega \) is given by the formula \( \omega = \sqrt{\frac{g}{L}} \), where \( g \) is the acceleration due to gravity (\( \approx 9.81 \, \text{m/s}^2 \)) and \( L \) is the length of the pendulum.
02

Calculate the natural frequency for each pendulum

We need to calculate \( \omega \) for each pendulum length using \( \omega = \sqrt{\frac{9.81}{L}} \). This will give us the natural frequency at which each pendulum oscillates.
03

Calculate natural frequency for pendulum (a)

For length \( L = 0.10 \, \text{m} \) (pendulum a), calculate \( \omega = \sqrt{\frac{9.81}{0.10}} \approx 9.90 \, \text{rad/s} \).
04

Calculate natural frequency for pendulum (b)

For length \( L = 0.30 \, \text{m} \) (pendulum b), calculate \( \omega = \sqrt{\frac{9.81}{0.30}} \approx 5.70 \, \text{rad/s} \).
05

Calculate natural frequency for pendulum (c)

For length \( L = 0.40 \, \text{m} \) (pendulum c), calculate \( \omega = \sqrt{\frac{9.81}{0.40}} \approx 4.95 \, \text{rad/s} \).
06

Calculate natural frequency for pendulum (d)

For length \( L = 0.80 \, \text{m} \) (pendulum d), calculate \( \omega = \sqrt{\frac{9.81}{0.80}} \approx 3.50 \, \text{rad/s} \).
07

Calculate natural frequency for pendulum (e)

For length \( L = 1.2 \, \text{m} \) (pendulum e), calculate \( \omega = \sqrt{\frac{9.81}{1.2}} \approx 2.86 \, \text{rad/s} \).
08

Calculate natural frequency for pendulum (f)

For length \( L = 2.8 \, \text{m} \) (pendulum f), calculate \( \omega = \sqrt{\frac{9.81}{2.8}} \approx 1.87 \, \text{rad/s} \).
09

Calculate natural frequency for pendulum (g)

For length \( L = 3.5 \, \text{m} \) (pendulum g), calculate \( \omega = \sqrt{\frac{9.81}{3.5}} \approx 1.67 \, \text{rad/s} \).
10

Calculate natural frequency for pendulum (h)

For length \( L = 5.0 \, \text{m} \) (pendulum h), calculate \( \omega = \sqrt{\frac{9.81}{5.0}} \approx 1.40 \, \text{rad/s} \).
11

Calculate natural frequency for pendulum (i)

For length \( L = 6.2 \, \text{m} \) (pendulum i), calculate \( \omega = \sqrt{\frac{9.81}{6.2}} \approx 1.25 \, \text{rad/s} \).
12

Identify pendulums within given frequency range

The beam oscillates between \(2.00 \) and \(4.00 \, \text{rad/s}\). Check which pendulums have natural frequencies within this range: Pendulums (d) and (e), with frequencies \(3.50 \, \text{rad/s} \) and \(2.86 \, \text{rad/s} \), fall within the given range.

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Key Concepts

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

Natural Frequency
The natural frequency of a system, like a simple pendulum, is the frequency at which the system naturally vibrates when not subjected to an external force or damping.
For a simple pendulum, the natural frequency is influenced by the length of the string or rod from which the mass is suspended.
The formula to calculate the angular natural frequency \( \omega \) for a pendulum is given by \( \omega = \sqrt{\frac{g}{L}} \), where \( g \) is the acceleration due to gravity (approximately \( 9.81\, \text{m/s}^2 \)) and \( L \) is the length of the pendulum.

The longer the pendulum, the lower its natural frequency, and vice versa:
  • Shorter pendulums will naturally oscillate faster.
  • Longer pendulums will oscillate more slowly.
This concept is crucial when determining which pendulums will oscillate strongly when an external force with a specific frequency is applied.
Oscillation
Oscillation refers to any repetitive variation, usually in time, of some measure around a central value or between two or more different states.
A simple pendulum oscillates back and forth due to the force of gravity acting on it.
When displaced from its rest position, the pendulum swings back and forth through an arc in a periodic manner.

Key characteristics of oscillation include:
  • A repeating motion that travels over the same path.
  • Each complete repetition is termed as a cycle.
  • The time it takes to complete one cycle is called the period.
  • The number of cycles per second is the frequency.
In a perfect setting without air resistance or friction, the amplitude (maximum displacement) of the pendulum's swing stays constant over time, allowing the pendulum to oscillate perpetually.
Angular Frequency
Angular frequency, commonly represented by \( \omega \), is a measure of how quickly an object goes through its oscillation cycle. It is linked closely to the concept of regular or linear frequency but is expressed in radians per second (\( \text{rad/s} \)).
For a simple pendulum, angular frequency tells how fast it oscillates back and forth.

Mathematically, angular frequency \( \omega \) is defined as: \[ \omega = 2\pi f \] where \( f \) is the linear frequency in hertz (\( \text{Hz} \)).
For simple harmonic motion like that of a pendulum, angular frequency also relates to the period \( T \) and is given by: \[ \omega = \frac{2\pi}{T} \]

Understanding angular frequency helps identify how the pendulum's length impacts the speed of oscillation:
  • High angular frequency means rapid oscillation.
  • Low angular frequency indicates slower oscillation.
Being mindful of angular frequency is vital in contexts where matching the oscillation speed to that of an external force—such as the beam in the original problem—is necessary for resonance.

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Most popular questions from this chapter

A particle executes linear SHM with frequency \(0.25 \mathrm{~Hz}\) about the point \(x=0\). At \(t=0\), it has displacement \(x=0.37 \mathrm{~cm}\) and zero velocity. For the motion, determine the (a) period, (b) angular frequency, (c) amplitude, (d) displacement \(x(t),(\mathrm{e})\) velocity \(v(t)\), (f) maximum speed, (g) magnitude of the maximum acceleration, (h) displacement at \(t=3.0 \mathrm{~s}\), and (i) speed at \(t=3.0 \mathrm{~s}\)

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 particle with a mass of \(1.00 \times 10^{-20} \mathrm{~kg}\) is oscillating with simple harmonic motion with a period of \(1.00 \times 10^{-5} \mathrm{~s}\) and a maximum speed of \(1.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\). Calculate (a) the angular frequency and (b) the maximum displacement of the particle.

A \(0.10 \mathrm{~kg}\) block oscillates back and forth along a straight line on a frictionless horizontal surface. Its displacement from the origin is given by $$ x=(10 \mathrm{~cm}) \cos [(10 \mathrm{rad} / \mathrm{s}) t+\pi / 2 \mathrm{rad}] $$ (a) What is the oscillation frequency? (b) What is the maximum speed acquired by the block? (c) At what value of \(x\) does this occur? (d) What is the magnitude of the maximum acceleration of the block? (e) At what value of \(x\) does this occur? (f) What force, applied to the block by the spring, results in the given oscillation?

What is the length of a simple pendulum whose full swing from left to right and then back again takes \(3.2 \mathrm{~s}\) ?

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