T = 2 pi sqrt(L/g)
1.4 = 6.28 sqrt (.47/g)
.47/g = .0497
g = 9.46 m/s^2
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
9 answers
It still says its the wrong answer. Let me calculate it again
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)
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)
I used .14 not .143
1.43 = 6.28 sqrt(.47/g)
try 9.06 m/s^2
try 9.06 m/s^2
For some reason its still saying its the wrong answer
Your saying g=9.06
yes
Yes, it finally worked thank you so much.
1 answer