QUESTION IMAGE
Question
current attempt in progress
in concept simulation 10.2 you can explore the concepts that are important in this problem. astronauts on a distant planet set up a simple pendulum of length 1.20 m. the pendulum executes simple harmonic motion and makes 100 complete oscillations in 400 s. what is the magnitude of the acceleration due to gravity on this planet?
g=
save for later
using multiple attempts will impact your score.
20% score reduction after attempt 2
attempts: 0 of 3 used submit answer
Step1: Calculate the period of the pendulum
The period \(T\) is the time per oscillation. Given \(n = 100\) oscillations in \(t=400\ s\), then \(T=\frac{t}{n}\).
\(T=\frac{400\ s}{100}=4\ s\)
Step2: Use the formula for the period of a simple pendulum
The formula for the period of a simple pendulum is \(T = 2\pi\sqrt{\frac{L}{g}}\), where \(L = 1.20\ m\) is the length of the pendulum and \(g\) is the acceleration due to gravity.
First, square both sides of the formula: \(T^{2}=4\pi^{2}\frac{L}{g}\).
Then, solve for \(g\): \(g = 4\pi^{2}\frac{L}{T^{2}}\)
Substitute \(L = 1.20\ m\) and \(T = 4\ s\) into the formula:
\(g=4\pi^{2}\times\frac{1.20}{4^{2}}\)
\(g=\pi^{2}\times\frac{1.20}{4}\)
\(g=\frac{9.8696\times1.20}{4}\)
\(g = 2.96\ m/s^{2}\)
Snap & solve any problem in the app
Get step-by-step solutions on Sovi AI
Photo-based solutions with guided steps
Explore more problems and detailed explanations
\(2.96\ m/s^{2}\)