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Chapter 12: Problem 60

A simple pendulum of length \(2.5 \mathrm{m}\) makes 5.0 complete swings in 16 s. What is the acceleration of gravity at the location of the pendulum?

Short Answer

Expert verified
The acceleration of gravity is approximately 9.71 m/s虏 at the location of the pendulum.

Step by step solution

01

Determine the Period of the Pendulum

The period (T) of a pendulum is the time it takes to complete one full swing. Given that it makes 5.0 swings in 16 seconds, we calculate the period as follows:\[ T = \frac{16}{5.0} = 3.2 \text{ seconds} \]
02

Recall the Formula for the Period of a Simple Pendulum

The formula for the period of a simple pendulum is given by:\[ T = 2\pi \sqrt{\frac{L}{g}} \]where \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.
03

Rearrange the Formula to Solve for Gravity

Rearrange the formula to solve for \( g \) (acceleration due to gravity):\[ g = \frac{4\pi^2L}{T^2} \]
04

Substitute the Known Values into the Formula

Using the formula from the previous step, substitute \( L = 2.5 \ m \) and \( T = 3.2 \ s \) into it:\[ g = \frac{4\pi^2 \times 2.5}{(3.2)^2} \]
05

Calculate the Numerical Value for Gravity

Perform the calculations to find \( g \):\[ g = \frac{4 \times 9.87 \times 2.5}{10.24} \approx 9.71 \, \mathrm{m/s^2} \]

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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 a fundamental concept in physics. It refers to how long it takes for a pendulum to swing once back and forth. This is crucial for understanding timekeeping and oscillatory motion. In this problem, we calculated that the pendulum takes 3.2 seconds to complete one full swing.
  • A full swing means going from one side to the other and back again.
  • Knowing the period helps in determining the pendulum's properties, such as the acceleration due to gravity.
To find the period, you need to divide the total time taken for multiple swings by the number of swings, as shown in the formula:\[ T = \frac{16}{5.0} = 3.2 \ \text{seconds} \]This formula gives you an idea of how the pendulum behaves over time. It's a stepping stone in further analysis.
Acceleration Due to Gravity
The acceleration due to gravity is a measure of how strongly gravity pulls objects towards the Earth. It varies slightly depending on where you are on the planet but is crucial for pendulum calculations.To find this from a pendulum, you first use the period and length of the pendulum with the formula:\[ T = 2\pi \sqrt{\frac{L}{g}} \]Rearranging to solve for \( g \):\[ g = \frac{4\pi^2L}{T^2} \]
  • Use the length \( L = 2.5 \ \text{m} \).
  • Use the period \( T = 3.2 \ \text{seconds} \).
Insert these into the formula, calculate, and you'll discover \( g \) is approximately \( 9.71 \, \mathrm{m/s^2} \). This calculated value is close to the standard Earth gravity, showing how pendulums help determine such forces.
Pendulum Length
The length of a pendulum is a key factor in determining its period, and it affects how fast or slow the pendulum will swing. A longer pendulum takes more time to complete a swing compared to a shorter one.In the context of a simple pendulum, the length is the distance from the pivot point where it's attached to the center of gravity of the pendulum mass.
  • The formula \( T = 2\pi \sqrt{\frac{L}{g}} \) relates length (\( L \)) to the period (\( T \)), showing the square root relationship.
  • When the length is known, it assists in calculating \( g \) when other factors are understood.
In our example with \( L = 2.5 \ \text{m} \), knowing this allows us to work backward to determine \( g \). Understanding pendulum length can also aid in clock design and various engineering applications where oscillation periods are crucial.

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Most popular questions from this chapter

A 0.321 -kg mass is attached to a spring with a force constant of \(13.3 \mathrm{N} / \mathrm{m}\). If the mass is displaced \(0.256 \mathrm{m}\) from equilibrium and released, what is its speed when it is \(0.128 \mathrm{m}\) from equilibrium?

The acceleration of a block attached to a spring is given by \(a=-\left(0.302 \mathrm{m} / \mathrm{s}^{2}\right) \cos ([2.41 \mathrm{rad} / \mathrm{s}] \mathrm{t})\) (a) What is the frequency of the block's motion? (b) What is the maximum speed of the block? (c) What is the amplitude of the block's motion?

An object undergoes simple harmonic motion of amplitude \(A\) and angular frequency \(\omega\) about the equilibrium point \(x=0\). Use energy conservation to show that the speed of the object at the general position \(x\) is given by the following expression: $$ v=\omega \sqrt{A^{2}-x^{2}} $$

A \(0.45-k g\) crow lands on a slender branch and bobs up and down with a period of 1.5 s. An eagle flies up to the same branch, scaring the crow away, and lands. The eagle now bobs up and down with a period of \(4.8 \mathrm{s}\). Treating the branch as an ideal spring, find (a) the effective force constant of the branch and (b) the mass of the eagle.

A mass on a spring oscillates with simple harmonic motion of amplitude \(A\) about the equilibrium position \(x=0 .\) Its maximum speed is \(v_{\max }\) and its maximum acceleration is \(a_{\mathrm{max}}\) (a) What is the speed of the mass at \(x=0 ?\) (b) What is the acceleration of the mass at \(x=0 ?\) (c) What is the speed of the mass at \(x=A ?\) (d) What is the acceleration of the mass at \(x=A ?\)
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