Question

in the pendulum formula ( t = 2pisqrt{\frac{l}{g}} ), we use ( g = 9.8m/s^{2} ) for the acceleration due to gravity on earth. but what about on the moon? if an astronaut on the surface of the moon swings a 1 - meter long pendulum, and it has a period of 4.9 seconds, what is the acceleration due to gravity, g, on the moon?
a. ( 0.7m/s^{2} )
b. ( 1.6m/s^{2} )
c. ( 2.8m/s^{2} )
d. ( 1.2m/s^{2} )

Explanation:

Step1: Rearrange the formula for \(g\)

Given \(T = 2\pi\sqrt{\frac{L}{g}}\), we can square both sides to get \(T^{2}=4\pi^{2}\frac{L}{g}\). Then, solve for \(g\): \(g = 4\pi^{2}\frac{L}{T^{2}}\).

Step2: Substitute the values of \(L\) and \(T\)

We know that \(L = 1m\) and \(T=4.9s\). Substitute these values into the formula for \(g\): \(g=\frac{4\pi^{2}\times1}{4.9^{2}}\).
Calculate \(4\pi^{2}\approx4\times9.8696 = 39.4784\) and \(4.9^{2}=24.01\).
Then \(g=\frac{39.4784}{24.01}\approx1.6m/s^{2}\)

Answer:

B. \(1.6m/s^{2}\)