Posted by Ray on Saturday, November 3, 2012 at 10:17pm.
Imagine you are an astronaut who has landed on another planet and wants to determine the freefall 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

Physics  Damon, Saturday, November 3, 2012 at 10:30pm
T = 2 pi sqrt(L/g)
1.4 = 6.28 sqrt (.47/g)
.47/g = .0497
g = 9.46 m/s^2

Physics  Ray, Saturday, November 3, 2012 at 10:37pm
It still says its the wrong answer. Let me calculate it again

Physics  Damon, Saturday, November 3, 2012 at 10:47pm
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)

Physics  Damon, Saturday, November 3, 2012 at 10:48pm
I used .14 not .143

Physics  Damon, Saturday, November 3, 2012 at 10:50pm
1.43 = 6.28 sqrt(.47/g)
try 9.06 m/s^2

Physics  Ray, Saturday, November 3, 2012 at 10:54pm
For some reason its still saying its the wrong answer

Physics  Ray, Saturday, November 3, 2012 at 10:56pm
Your saying g=9.06

Physics  Damon, Saturday, November 3, 2012 at 10:57pm
yes

Physics  Raven, Sunday, November 4, 2012 at 2:25pm
Yes, it finally worked thank you so much.
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