Question

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 300 s. what is the magnitude of the acceleration due to gravity on this planet? g=

Explanation:

Step1: Calculate the period \( T \)

The period \( T \) is the time for one complete oscillation. Given \( N = 100 \) oscillations in \( t=300\ s \), then \( T=\frac{t}{N} \).
\( T=\frac{300\ s}{100}=3\ s \)

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 \( g \): \( g=\frac{4\pi^{2}L}{T^{2}} \)

Step3: Substitute the values of \( L \) and \( T \)

Substitute \( L = 1.20\ m \) and \( T = 3\ s \) into the formula for \( g \):
\( g=\frac{4\pi^{2}\times1.20}{3^{2}} \)
\( g=\frac{4\times3.14^{2}\times1.20}{9} \)
\( g=\frac{4\times9.86\times1.20}{9} \)
\( g=\frac{47.328}{9}\approx5.26\ m/s^{2} \)

Answer:

\( 5.26\ m/s^{2} \)