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Chapter 13: Problem 81

A simple pendulum is set into small-angle motion, making a maximum angle with the vertical of \(5^{\circ} .\) Its period is 2.21 s. (a) Determine its length. (b) Determine its maximum speed. (c) What is the acceleration of the pendulum bob when it is at the lowest position?

Short Answer

Expert verified
Length is 1.21 m, max speed is 0.30 m/s, and acceleration at lowest position is 9.74 m/s虏.

Step by step solution

01

Understand the Period of a Simple Pendulum

The period of a simple pendulum is described by the formula: \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( T \) is the period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity (approximately \( 9.81 \ m/s^2 \)).
02

Solve for Length of the Pendulum

Rearrange the period formula to solve for \( L \): \( L = \left(\frac{T}{2\pi}\right)^2 g \). Substitute \( T = 2.21 \ s \) and \( g = 9.81 \ m/s^2 \):\[ L = \left(\frac{2.21}{2\pi}\right)^2 \times 9.81 \approx 1.21 \ m \].
03

Calculate Maximum Speed of the Pendulum Bob

The maximum speed \( v_{max} \) occurs at the lowest point, given by the formula: \( v_{max} = \omega L \theta_{max} \), where \( \omega = \frac{2\pi}{T} \) is the angular frequency, \( L \) is the length, and \( \theta_{max} \) is the maximum angle in radians. Convert \( \theta = 5^{\circ} \) to radians: \( \theta = \frac{5\pi}{180} \approx 0.0873 \ rad \). Substitute \( L = 1.21 \ m \), \( T = 2.21 \ s \):\[ \omega = \frac{2\pi}{2.21} \approx 2.84 \ rad/s \]\[ v_{max} = 2.84 \times 1.21 \times 0.0873 \approx 0.30 \ m/s \].
04

Determine Acceleration at Lowest Position

At the lowest position, the centripetal acceleration is considered. It can be calculated using \( a = \omega^2 L \). Using previous calculations of \( \omega \) and \( L \):\[ a = (2.84)^2 \times 1.21 \approx 9.74 \ m/s^2 \].

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Key Concepts

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

Pendulum Period
The period of a simple pendulum is the time it takes for the pendulum to complete one full back and forth swing. You can find this period using the formula:
  • \( T = 2\pi \sqrt{\frac{L}{g}} \)
In the formula:
  • \( T \) is the period (in seconds).
  • \( L \) is the length of the pendulum (in meters).
  • \( g \) is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \)).
The period is directly influenced by the length of the pendulum and gravity. Longer pendulums have longer periods. If you measure how long it takes for your pendulum to swing back and forth, you can estimate its length by rearranging this handy equation. This is useful for understanding how changes in gravity, like on different planets, would change the swing time.
Pendulum Length Calculation
To calculate the length of a pendulum when given its period, we rearrange the period formula to solve for \( L \):
  • \( L = \left(\frac{T}{2\pi}\right)^2 g \)
Plugging in the values for our problem:
  • \( T = 2.21 \, s \)
  • \( g = 9.81 \, m/s^2 \)
We compute:
  • \( L = \left(\frac{2.21}{2\pi}\right)^2 \times 9.81 \approx 1.21 \, m \)
This gives us the length of the pendulum. Essentially, you measure the swing time and apply this formula to find the physical length needed for that period. It's a straightforward calculation that gives insight into not just this pendulum but any system following similar harmonic motion.
Maximum Speed of Pendulum Bob
The maximum speed of a pendulum bob is found at its lowest point in the swing, where its energy transformation from potential to kinetic is greatest. To find the maximum speed, use the formula:
  • \( v_{max} = \omega L \theta_{max} \)
Where:
  • \( \omega = \frac{2\pi}{T} \) is the angular frequency.
  • \( \theta_{max} \) is the maximum angle, which needs to be in radians.
Convert the angle from degrees to radians:
  • \( \theta = \frac{5\pi}{180} \approx 0.0873 \) radians
Then, substitute the known values:
  • \( L = 1.21 \, m \)
  • \( \omega \approx 2.84 \, rad/s \)
Thus:
  • \( v_{max} = 2.84 \times 1.21 \times 0.0873 \approx 0.30 \, m/s \)
This speed represents the swiftest movement of the bob as it swings, controlled by its length and the angle of release.
Pendulum Acceleration
At its lowest point, the pendulum bob also experiences the greatest centripetal acceleration. This acceleration in circular motion can be calculated using the formula:
  • \( a = \omega^2 L \)
Given:
  • \( \omega = 2.84 \, rad/s \)
  • \( L = 1.21 \, m \)
We calculate:
  • \( a = (2.84)^2 \times 1.21 \approx 9.74 \, m/s^2 \)
This shows how strongly the bob is being pulled towards the center of its path. It's important to note that this acceleration is due to the pendulum鈥檚 motion, rather than gravity alone. Understanding these forces explains why pendulums behave the way they do, which is crucial for both theoretical studies and practical applications such as clocks.

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Most popular questions from this chapter

A 0.500 -kg mass is attached to a vertical spring and the system is allowed to come to equilibrium. The mass is then given an initial downward speed of \(1.50 \mathrm{~m} / \mathrm{s} .\) The mass travels downward \(25.3 \mathrm{~cm}\) before stopping and returning. (a) Determine the spring constant. (b) What is its speed after it falls \(15.0 \mathrm{~cm} ?\) (c) What is the acceleration of the mass at the very bottom of the motion?

\(\bullet\bullet\) The equation of motion of a SHM oscillator is \(x=(0.50 \mathrm{~m}) \sin (2 \pi f) t,\) where \(x\) is in meters and \(t\) is in seconds. If the position of the oscillator is at \(x=0.25 \mathrm{~m}\) at \(t=0.25 \mathrm{~s},\) what is the frequency of the oscillator?

\(\bullet\) A particle oscillates in SHM with an amplitude \(A\). What is the total distance (in terms of \(A\) ) the particle travels in three periods?

\(\bullet\) The displacement of an object is given by \(y=(5.0 \mathrm{~cm}) \cos [(20 \pi \mathrm{rad} / \mathrm{s}) t] .\) What are the object's (a) amplitude, (b) frequency, and (c) period of oscillation?

\(\bullet\bullet\bullet\) The forces acting on a simple pendulum are shown in \(\nabla\) Fig. \(13.26 .\) (a) Show that, for the small angle approximation \((\sin \theta \approx \theta),\) the force producing the motion has the same form as Hooke's law. (b) Show by analogy with a mass on a spring that the period of a simple pendulum is given by \(T=2 \pi \sqrt{L / g}\). [Hint: Think of the effective spring constant.]
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