Question

A pendulum of length of 1 m oscillates near the surface of Planet X. What is the surface gravity of the planet if the period of oscillation is 4 s?
A. 1.6 ms-2
B. 2.5 ms-2
C. 6.4 ms-2
D. 9.8 ms-2

Answer 1

\[T=2\pi \sqrt{\frac{ l }{ g }} \]

you can use this formula..

Answer 2

Can you show that formula with the terms plugged in? I got like 7.6ms-3 It was really weird.

Answer 3

take a square of both side
\[T ^{2} = 4\pi ^{2} \frac{ l }{ g } \]
then
\[g=4\pi ^{2}\frac{ l }{ T ^{2} } \]

where ;

T=4 s , l=1 m

\[g=4*(3,14)^{2}\frac{ 1 }{ 16 } \]

g= 2.45 m/s^2

Answer 4

Thank you! I see my problem. I didnt square both sides. Thanks again!

Answer 5

you are welcome!

Answer 6

this formula only holds for small angles be carefull