Imagine you are an astronaut who has landed on another planet and wants to determine the free-fall acceleration on that planet. In one of the experiments you decide to conduct, you use a pendulum \(0.50 \mathrm{~m}\) long and find that the period of oscillation for this pendulum is \(1.50 \mathrm{~s}\). What is the acceleration due to gravity on that planet?

Short Answer

Expert verified
Answer: The acceleration due to gravity on this planet is approximately \(6.076 \mathrm{~m/s^2}\).

Step by step solution

01

Write down the known values

We know the length of the pendulum, L = 0.50 m and the period of oscillation, T = 1.50 s.
02

Write the formula for the period of oscillation

The formula for the period of oscillation of a pendulum is given by \(T=2\pi\sqrt{\frac{L}{g}}\).
03

Plug the known values into the formula

We know the values of T and L, so our equation becomes: \(1.50 = 2\pi\sqrt{\frac{0.50}{g}}\)
04

Solve for the acceleration due to gravity, g

To solve for g, we need to isolate g in the equation. First, we can divide both sides of the equation by 2\(\pi\), which results in: \(\frac{1.50}{2\pi}=\sqrt{\frac{0.50}{g}}\) Next, we square both sides of the equation to eliminate the square root: \(\left(\frac{1.50}{2\pi}\right)^2 = \frac{0.50}{g}\) Now, we can cross multiply to isolate g: \(g = \frac{0.50}{\left(\frac{1.50}{2\pi}\right)^2}\)
05

Compute the value of g

Compute the value of g using a calculator: \(g = \frac{0.50}{\left(\frac{1.50}{2\pi}\right)^2} \approx 6.076 \mathrm{~m/s^2}\)
06

State the final answer

The acceleration due to gravity on this planet is approximately \(6.076 \mathrm{~m/s^2}\).

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