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Astronauts on a distant planet set up a simple pendulum of length 1.2 \(\mathrm{m}\) . The pendulum executes simple harmonic motion and makes 100 complete vibrations in 280 s. What is the magnitude of the acceleration due to gravity on this planet?

Short Answer

Expert verified
The acceleration due to gravity on the planet is approximately 6.01 m/s².

Step by step solution

01

Understand the Problem

We need to find the magnitude of the acceleration due to gravity on a distant planet using a pendulum's motion. We are given the pendulum length (\( L = 1.2 \, \text{m} \)) and the time for 100 vibrations (\( 280 \, \text{s} \)).
02

Calculate the Period of the Pendulum

The period \( T \) of the pendulum is the time taken for one complete vibration. Since the pendulum makes 100 vibrations in 280 seconds, the period \( T \) can be calculated using the formula: \[ T = \frac{\text{Total Time}}{\text{Number of Vibrations}} = \frac{280}{100} = 2.8 \, \text{s} \]
03

Use the Formula for a Pendulum's Period

The formula for the period \( T \) of a simple pendulum is: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \( L \) is the length and \( g \) is the acceleration due to gravity. We need to solve for \( g \).
04

Solve for Acceleration Due to Gravity

To find \( g \), we first rearrange the formula from step 3: \[ g = \frac{4\pi^2 L}{T^2} \] Substitute the known values: \[ g = \frac{4 \times \pi^2 \times 1.2}{2.8^2} \] Now, calculate \( g \).
05

Calculate the Value of g

Plug the numbers into a calculator: \[ g = \frac{4 \times 9.8696 \times 1.2}{7.84} \approx \frac{47.1224}{7.84} \approx 6.01 \] The value of \( g \) is approximately \( 6.01 \, \text{m/s}^2 \).

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Key Concepts

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

Simple Pendulum
A simple pendulum is a basic mechanical system consisting of a suspended weight hung from a fixed point. The arrangement typically involves a mass (called the "bob") attached to the end of a lightweight string or rod. When displaced from its equilibrium position and released, the pendulum swings back and forth. This motion occurs due to the force of gravity acting on the bob.

Key characteristics of a simple pendulum include:
  • Bob: The weight at the end of the string.
  • Length: The distance from the pivot point to the center of the bob.
  • Pivot Point: The fixed point from which the pendulum is suspended.
  • Amplitude: The maximum extent of the swing on either side of the equilibrium.
Many factors can affect the pendulum's behavior, including the length of the string and the gravitational force acting upon it. However, other factors like the mass of the bob or the amplitude of the swing do not affect the time period of the swing significantly.
Acceleration Due to Gravity
Acceleration due to gravity, denoted as \( g \), is a crucial concept in physics as it defines how quickly an object accelerates under the influence of gravitational force. On Earth, the standard value of \( g \) is approximately 9.81 \( m/s^2 \). However, the value of \( g \) may vary on different celestial bodies, depending on their mass and radius.

In the example of the distant planet, we calculated \( g \) to be approximately 6.01 \( m/s^2 \). This indicates that the planet has either less mass or a larger radius compared to Earth, resulting in a weaker gravitational pull.

The formula used in pendulum experiments to determine \( g \) involves measuring the period of oscillation. By knowing the pendulum's length and the time period, \( g \) can be calculated accurately, providing valuable information about the planet's gravitational characteristics.
Simple Harmonic Motion
Simple harmonic motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement but in the opposite direction. For a pendulum, SHM is observed as it sways back and forth in a rhythmic pattern.

In the context of a pendulum, SHM can be described with the following points:
  • Restoring Force: The force that brings the pendulum back to its equilibrium position.
  • Displacement: The distance the pendulum moves from its rest position.
  • Frequency: The number of complete cycles the pendulum makes in a given time.
  • Period: The time it takes to complete one full cycle.
The pendulum fits the criteria for SHM because the gravitational force acting on it is consistent. This ensures the motion remains regular, making it predictable and easily analyzable. Factors such as damping and external forces can, however, influence this ideal motion, but in our simplified context, we often ignore them to focus on core principles.
Pendulum Period Calculation
Understanding how to calculate the period of a simple pendulum is key for determining the gravitational acceleration of a particular environment. The period \( T \) represents the time it takes for the pendulum to swing back and forth once fully.

The formula used for calculating the period of a simple pendulum is derived from physics principles:\[ T = 2\pi \sqrt{\frac{L}{g}} \]Where:
  • \( T \) is the period of the pendulum.
  • \( L \) is the length of the pendulum.
  • \( g \) is the acceleration due to gravity.
To find \( T \) in our example, we divide the total time for a set number of vibrations by the number of vibrations, yielding 2.8 seconds per vibration. By rearranging the formula to solve for \( g \), we have:\[ g = \frac{4\pi^2 L}{T^2} \]Using this rearranged equation, you can pinpoint the value of \( g \) by substituting known values such as \( L \) and \( T \). This process allows researchers and physicists to test and explore gravitational forces on various planets merely through pendulum-like systems.

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Most popular questions from this chapter

An 11.2 -kg block and a 21.7 -kg block are resting on a horizontal frictionless surface. Between the two is squeezed a spring (spring constant \(=1330 \mathrm{N} / \mathrm{m}\) ). The spring is compresed by 0.141 \(\mathrm{m}\) from its unstrained length and is not attached to either block. With what speed does each block move away after the mechanism keeping the spring squeezed is released and the spring falls away?

A 1.1\(\cdot \mathrm{kg}\) object is suspended from a vertical spring whose spring constant is 120 \(\mathrm{N} / \mathrm{m}\) . (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of 0.20 \(\mathrm{m}\) and released from rest. Find the speed with which the object passes through its original position on the way up.

A hand exerciser utilizes a coiled spring. A force of 89.0 \(\mathrm{N}\) is required to compress the spring by 0.0191 \(\mathrm{m} .\) Determine the force needed to compress the spring by 0.0508 \(\mathrm{m} .\)

A spiral staircase winds up to the top of a tower in an old castle. To measure the height of the tower, a rope is attached to the top of the tower and hung down the center of the staircase. However, nothing is available with which to measure the length of the rope. Therefore, at the bottom of the rope a small object is attached so as to form a simple pendulum that just clears the floor. The period of the pendulum is measured to be 9.2 s. What is the height of the tower?

The length of a simple pendulum is 0.79 \(\mathrm{m}\) and the mass of the particle (the bob) at the end of the cable is 0.24 \(\mathrm{kg}\) . The pendulum is pulled away from its equilibrium position by an angle of \(8.50^{\circ}\) and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing?

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