Asked by Ray


Imagine you are an astronaut who has landed on another planet and wants to determine the free-fall acceleration on that planet. In one of the experiments you decide to conduct, you use a pendulum 0.47 m long and find that the period of oscillation for this pendulum is 1.43 s. What is the acceleration due to gravity on that planet? (Enter the magnitude only.)

I know the formula to use is
T=2pi(L/g)^0.5(g) but I keep getting a wrong answer

Answers

Answered by Damon
T = 2 pi sqrt(L/g)
1.4 = 6.28 sqrt (.47/g)
.47/g = .0497
g = 9.46 m/s^2
Answered by Ray
It still says its the wrong answer. Let me calculate it again
Answered by Damon
That assumes that you know about simple pendula

small angle A at top, g and L and m
KE at bottom = (1/2) m V^2
V is max speed at bottom = L d angle/dt
Pe at top = m g h = m g L(1- cos A)
for small A, cos A = 1 - A^2/2 + ....

m g L (A^2/2) = (1/2) m V^2

g L A^2 = V^2
now assume sinusoidal motion
angle = A sin (2 pi/T) t
d angle/dt = (2 pi/T) A cos (2 pi/T)t
x max = A
v max = L d angle/dt = (2 pi L/T)A
then
g L A^2 = (2 pi L/T)^2 A^2

g/L = (2 pi/T)^2
T^2 = (2 pi)^2 L/g
T = 2 pi sqrt (L/g)
Answered by Damon
I used .14 not .143
Answered by Damon
1.43 = 6.28 sqrt(.47/g)
try 9.06 m/s^2
Answered by Ray
For some reason its still saying its the wrong answer
Answered by Ray
Your saying g=9.06
Answered by Damon
yes
Answered by Raven
Yes, it finally worked thank you so much.
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions