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

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

anonymous · Answer

$$T=2\pi \sqrt{\frac{ l }{ g }}  $$  

you can use this formula..

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

you are welcome!

anonymous · Answer

this formula only holds for small angles be carefull