Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 cm. She finds that the pendulum makes 100 complete swings in 136 s. What is the value of \(g\) on this planet?

Short answer

The acceleration due to gravity, \( g \), is approximately 10.67 m/s² on that planet.

Step-by-step solution

Understand the Pendulum Formula

The period of a simple pendulum, which is the time for one complete swing back and forth, is given by the formula \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \( T \) is the period in seconds, \( L \) is the length of the pendulum in meters, and \( g \) is the acceleration due to gravity.

Find the Period of One Swing

The period \( T \) is the total time for the pendulum to make one complete swing. The total time for 100 swings is 136 seconds. Therefore, the period for one swing is: \[ T = \frac{136 \text{ s}}{100} = 1.36 \text{ s} \]

Convert Length to Meters

The length of the pendulum is 50.0 cm. Convert this into meters by dividing by 100:\[ L = \frac{50.0}{100} = 0.50 \text{ m} \]

Rearrange Formula to Solve for g

We need to solve for \( g \) in the pendulum formula. Rearrange the formula to:\[ g = \frac{4\pi^2L}{T^2} \]

Substitute Values and Calculate g

Now, substitute \( L = 0.50 \text{ m} \) and \( T = 1.36 \text{ s} \) into the rearranged formula to calculate \( g \):\[ g = \frac{4\pi^2 (0.50)}{(1.36)^2} \]Calculate the result:1. \( 1.36^2 = 1.8496 \)2. \( 4\pi^2 = 39.4784 \) (since \( \pi^2 \approx 9.8696 \))3. Therefore, \( \frac{39.4784 \times 0.50}{1.8496} \equiv \frac{19.7392}{1.8496} \approx 10.67 \text{ m/s}^2 \)

Key concepts

Acceleration Due to Gravity

Acceleration due to gravity, often symbolized as \( g \), is a key factor in understanding how objects fall or oscillate. It represents the rate at which an object accelerates when it is in free fall. On Earth, this value is approximately 9.81 m/s², but it can differ on other planets depending on their masses and radii.
In pendulum physics, \( g \) can be calculated using the pendulum's period and length. The acceleration due to gravity affects how swiftly a pendulum swings back and forth. This is essential in determining time periods in various planetary environments, as seen in our exercise where all other variables aside from the pendulum's length and time period are a mystery.

• Gravity differs from planet to planet.
• It determines the speed of oscillation for a pendulum.
• Knowing \( g \) is crucial for astronomical calculations.

Pendulum Period

The pendulum period is the time it takes for a pendulum to make one full oscillation, swinging forward and then back to the starting point. The period (\( T \)) is linked with the pendulum's length (\( L \)) and the acceleration due to gravity (\( g \)).
In our exercise, the space explorer determines the period of the pendulum's swing by timing how long it takes for the pendulum to complete 100 swings – a total of 136 seconds. This measurement is then averaged to find a single swing time: 1.36 seconds.

• The period helps in determining gravitational forces on planets.
• It's derived by dividing total swing time by the number of swings.
• A longer period generally suggests weaker gravity if \( L \) remains constant.

Simple Pendulum Formula

The simple pendulum formula is crucial in analyzing the motion of pendulums. It ties together the pendulum's period, length, and the gravity it experiences.\[T = 2\pi \sqrt{\frac{L}{g}}\]This formula describes how the three elements interact: \( T \) (period) is the time for a complete swing, \( L \) (length) is measured in meters, and \( g \) (acceleration due to gravity) affects everything.
To isolate \( g \), we rearrange the formula:\[g = \frac{4\pi^2L}{T^2}\]By plugging in the known values of period and length, we can accurately find \( g \) on an unknown planet, as we did in the exercise.

• The formula showcases the balance between length, period, and acceleration due to gravity.
• Rearranging the formula lets us solve for unknown gravitational forces.
• It is an essential tool for physics, especially in environments with unknown gravitational conditions.