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 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

- 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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