Problem 47 - College Physics

Question

\(\bullet\) \(\bullet\) A pendulum on Mars. A certain simple pendulum has a period on earth of 1.60 s. What is its period on the surface of Mars, where the acceleration due to gravity is 3.71 \(\mathrm{m} / \mathrm{s}^{2}\) ?

Step-by-Step Solution

Answer

The period of the pendulum on Mars is approximately 2.63 seconds.

Step 1: Understanding the Formula for Pendulum Period

The period of a simple pendulum is given 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. Our goal is to find out how the period changes when the pendulum is taken to Mars.

Step 2: Identify Known Values on Earth

On Earth, we know the period is \( T_{\text{Earth}} = 1.60 \) seconds. The acceleration due to gravity on Earth, \( g_{\text{Earth}}, is 9.81 \, \text{m/s}^2 \). These will be used to find the length of the pendulum.

Step 3: Calculate the Length of the Pendulum

Since \( T_{\text{Earth}} = 2\pi \sqrt{\frac{L}{g_{\text{Earth}}}} \), we can rearrange to find \( L \): \( L = \frac{T_{\text{Earth}}^2 \cdot g_{\text{Earth}}}{4\pi^2} \). Substituting \( T_{\text{Earth}} = 1.60 \) and \( g_{\text{Earth}} = 9.81 \), we calculate \( L \).

Step 4: Calculate the Period on Mars

Using the expression \( T_{\text{Mars}} = 2\pi \sqrt{\frac{L}{g_{\text{Mars}}}} \), and knowing \( g_{\text{Mars}} = 3.71 \), we substitute the previously found value of \( L \) to find \( T_{\text{Mars}} \).

Step 5: Solve for the Period on Mars

After substituting the values into the formula, calculate \( T_{\text{Mars}} \). This will give the pendulum's period on Mars.

Key Concepts

Simple Pendulum

A simple pendulum is a weight suspended from a pivot, such that it can swing back and forth. This setup is often used to model periodic motion and involves just a few key components: the mass of the weight (often called the bob), the length of the string or rod it hangs from, and the effect of gravity on the system.

• In ideal conditions, the only force acting on the pendulum is gravity.
• The motion is periodic, meaning it repeats in a regular time interval called the period.

The period of a simple pendulum, which is the time it takes for the pendulum to complete one full swing back and forth, is determined not by the mass but by the length of the pendulum and the acceleration due to gravity.
The basic formula for the period of a simple pendulum is given by \[ T = 2\pi \sqrt{\frac{L}{g}} \] where:

• \( T \) is the period of the pendulum.
• \( L \) is the length of the pendulum.
• \( g \) is the acceleration due to gravity.

This means that to change the period of a simple pendulum, you must modify either the length of the pendulum or the gravitational acceleration acting upon it.

Acceleration Due to Gravity

Acceleration due to gravity, denoted as \( g \), is the acceleration experienced by an object alone in a gravitational field, typically near the surface of a planet. On Earth, this standard acceleration is approximately 9.81 \, \text{m/s}^2.

• This value can vary slightly depending on altitude and geographical location due to Earth's non-uniform shape and density.
• On other planets or celestial bodies, or in space, \( g \) can differ significantly.

The concept of gravity plays a crucial role in pendulum motion as it affects how quickly the pendulum swings. A larger gravitational pull will result in a shorter period because the pendulum swings faster. Conversely, a smaller gravitational force, such as on Mars, will cause the pendulum to have a longer period, as it swings more slowly.
On Mars, the gravity is much weaker, measured at about 3.71 \, \text{m/s}^2. This lower acceleration due to gravity means any pendulum on Mars will have a longer period than the same pendulum on Earth.

Period on Mars

To find the period of a pendulum on Mars, it's essential to understand how the change in gravitational acceleration affects it. Given that the period of a pendulum depends on gravity, using Mars' gravity of 3.71 \, \text{m/s}^2 leads to a noticeable change in the pendulum's period compared to Earth.
By applying the pendulum period formula \[ T = 2\pi \sqrt{\frac{L}{g}} \], and substituting Mars' gravity into it, you calculate the pendulum's altered period on Mars:

• Determine the length of the pendulum from its known period on Earth using Earth's gravity.
• Use this length with Mars' gravity to find the new period on Mars.

Specifically, since Mars has approximately \( 3.71\, \text{m/s}^2 \) compared to Earth's \( 9.81 \, \text{m/s}^2 \), the pendulum’s period would be longer due to the reduced gravitational pull. This offers an insightful demonstration of how gravity affects motion differently on various planets, emphasizing the importance of understanding physical laws beyond Earth’s boundaries.