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Chapter 15: Problem 91

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 Hz, (b) 0.388 Hz, (c) 0 Hz.

Step by step solution

01

Formula for Frequency

The frequency of a simple pendulum is given by the formula: \( f = \frac{1}{2\pi} \sqrt{\frac{g}{L}} \), where \( f \) is the frequency, \( g \) is the acceleration due to gravity, and \( L \) is the length of the pendulum. In this problem, \( L = 2.0 \ m \).
02

Calculate Frequency in a Room

In a room, the acceleration due to gravity, \( g \), is approximately \( 9.8 \ m/s^2 \). So, substituting into the formula: \( f = \frac{1}{2\pi} \sqrt{\frac{9.8}{2.0}} \). Simplifying, we get \( f \approx 0.353 \ Hz \).
03

Calculate Frequency in an Elevator Accelerating Upward

In an elevator accelerating upward, the effective acceleration due to gravity becomes \( g + a \), where \( a = 2.0 \ m/s^2 \). Therefore, the effective gravity is \( 11.8 \ m/s^2 \). Substituting into the formula gives: \( f = \frac{1}{2\pi} \sqrt{\frac{11.8}{2.0}} \). Simplifying, we get \( f \approx 0.388 \ Hz \).
04

Calculate Frequency in Free Fall

In free fall, the pendulum and the elevator both fall freely, resulting in zero effective gravity \( g_{eff} = 0 \). Therefore, the formula \( f = \frac{1}{2\pi} \sqrt{\frac{g_{eff}}{L}} \) yields \( f = 0 \) as there is no restoring force to cause oscillation.

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Key Concepts

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

Frequency Calculation
Understanding the frequency of a simple pendulum is crucial to grasping how it oscillates. The frequency (\(f\)) is simply the number of times the pendulum swings back and forth in a second. For a pendulum, the formula to calculate its frequency is \(f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}\).
This equation relates the frequency to two important factors:
  • The length of the pendulum (\(L\)). A longer pendulum has a larger "swing" and thus a lower frequency.
  • The acceleration due to gravity (\(g\)). This is the force that pulls the pendulum downward. Greater gravity increases the frequency as it accelerates the pendulum's motion.
Calculating the pendulum's frequency will involve substituting known values for \(g\) and \(L\). This offers insight into how pendulums behave differently depending on their environment.
Acceleration Due to Gravity
Acceleration due to gravity (\(g\)) is a key player in the pendulum's oscillating behavior. On Earth, this is typically about 9.8 \(m/s^2\), a measure of how fast objects accelerate when falling freely.
Gravity affects pendulums in a straightforward yet profound way:
  • A higher gravitational pull results in a faster oscillation frequency as it pulls the pendulum more strongly.
  • Conversely, if gravity is weaker, the frequency will decrease, resulting in slower swings.
Throughout different environments or setups, the effective gravity may change, and it will directly impact the pendulum's movement. Imagining how gravity influences everyday simple pendulums helps form the basis of more complex scenarios in classical mechanics.
Effect of Acceleration on Pendulum
When a pendulum is in an accelerating system, such as an elevator moving upward, the effective gravitational force acting on the pendulum changes. This concept shows how motion in one direction can impact vertical gravitational effects:
  • If the elevator accelerates upwards, the effective gravity becomes the sum of local gravity and the elevator's acceleration, given by \(g + a\). This creates an enhanced pull on the pendulum.
  • This increased effective gravity results in a higher frequency, making the pendulum oscillate more quickly than it would under standard gravity alone.
Understanding how different accelerations interact with basic gravitational forces helps to see how pendulums may behave differently in non-static scenarios, leading to a more comprehensive understanding of dynamics in physics.
Free Fall Dynamics
In free fall, a pendulum and an elevator fall together, leading to unique dynamics. Here, the effective gravitational acceleration is zero, creating an environment with no net gravitational force acting on the pendulum.
Key points to consider on why this matter:
  • The absence of gravity means the restoring force that causes pendulum swings is non-existent, causing the pendulum to cease oscillation.
  • Thus, the frequency in free fall equals zero because there is no force to "activate" its swinging motion.
This situation offers insight into how pivotal gravity is for pendulums and underscores gravity's importance in many physical phenomena.

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Most popular questions from this chapter

Hanging from a horizontal beam are nine simple pendulums of the following lengths: \((\mathrm{a}) 0.10,(\mathrm{~b}) 0.30,(\mathrm{c}) 0.40,(\mathrm{~d}) 0.80,(\mathrm{e}) 1.2,(\mathrm{f}) 2.8,\) (g) \(3.5,\) (h) 5.0 , and (i) 6.2 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?

A thin uniform rod \((\mathrm{mass}=0.50 \mathrm{~kg})\) swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. The rod swings with a period of \(1.5 \mathrm{~s}\) and an angular amplitude of \(10^{\circ} .\) (a) What is the length of the rod? (b) What is the maximum kinetic energy of the rod as it swings?

A 50.0 g stone is attached to the bottom of a vertical spring and set vibrating. If the maximum and the period is \(0.500 \mathrm{~s}\), find the (a) spring constant of the spring, (b) amplitude of the motion, and (c) frequency of oscillation.

A \(4.00 \mathrm{~kg}\) block hangs from a spring, extending it \(16.0 \mathrm{~cm}\) from its unstretched position. (a) What is the spring constant? (b) The block is removed, and a \(0.500 \mathrm{~kg}\) body is hung from the same spring. If the spring is then stretched and released, what is its period of oscillation?

A damped harmonic oscillator consists of a block \((m=2.00 \mathrm{~kg}),\) a spring \((k=10.0 \mathrm{~N} / \mathrm{m}),\) and a damping force \((F=-b v) .\) Initially, it oscillates with an amplitude of \(25.0 \mathrm{~cm}\) because of the damping, the amplitude falls to three-fourths of this initial value at the completion of four oscillations. (a) What is the value of \(b ?\) (b) How much energy has been "lost" during these four oscillations?
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