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
T = 2 pi sqrt(L/g)
1.4 = 6.28 sqrt (.47/g)
.47/g = .0497
g = 9.46 m/s^2
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)
I used .14 not .143
1.43 = 6.28 sqrt(.47/g)
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.
To determine the acceleration due to gravity on the planet, you can use the formula you mentioned, but you need to rearrange it slightly to solve for the acceleration.
The formula is:
T = 2π * √(L/g)
Where:
T = Period of oscillation (1.43 s)
L = Length of the pendulum (0.47 m)
g = Acceleration due to gravity (which we need to find)
First, square both sides of the equation to isolate g:
(T/2π)^2 = L/g
Next, rearrange the equation to solve for g:
g = L / (T/2π)^2
Now let's substitute the given values into the equation:
g = 0.47 / (1.43/2π)^2
To calculate that value, we need to know the value of π. Assuming we use the approximation 3.14 for π, we can proceed with the calculation:
g = 0.47 / (1.43/2*3.14)^2
g = 0.47 / (0.7096)^2
g ≈ 0.47 / 0.5027
g ≈ 0.935 m/s^2
So, the approximate acceleration due to gravity on that planet is 0.935 m/s^2.