In Concept Simulation 10.2 you can explore the concepts that are important in this problem. Astronauts on a distant planet set up a simple pendulum of length 1.2 m. The pendulum executes simple harmonic motion and makes 100 complete oscillations in 330 s. What is the magnitude of the acceleration due to gravity on this planet?
To find the magnitude of the acceleration due to gravity on the planet, we can use the equation for the period of a simple pendulum:
T = 2π√(L/g)
where:
T is the period of the pendulum,
L is the length of the pendulum, and
g is the acceleration due to gravity.
In this problem, we are given the length of the pendulum (L = 1.2 m) and the number of complete oscillations it makes in a given time (100 oscillations in 330 s), which allows us to calculate the period of the pendulum (T).
To find the period (T), we can use the formula:
T = (time taken for 100 oscillations) / (number of oscillations)
T = 330 s / 100
Now, we can substitute the given values into the formula for the period of the pendulum:
T = 2π√(L/g)
330 s / 100 = 2π√(1.2 m / g)
Rearranging the formula, we get:
(330 s / 100)^2 = 2π^2 * (1.2 m / g)
Now, we can solve for g:
g = 2π^2 * (1.2 m) / (330 s / 100)^2
Calculating the value, we find:
g ≈ 9.8 m/s^2
Therefore, the magnitude of the acceleration due to gravity on this planet is approximately 9.8 m/s^2.
To find the magnitude of the acceleration due to gravity on the planet, we can use the equation for the period of a simple pendulum:
T = 2π√(L/g)
Where T is the period of the pendulum, L is the length of the pendulum, and g is the acceleration due to gravity.
In this case, we are given the period (330 s) and the length (1.2 m) of the pendulum. We need to rearrange the equation to solve for g:
T = 2π√(L/g)
Squaring both sides of the equation gives:
T^2 = 4π^2(L/g)
Now we can solve for g by isolating it on one side of the equation:
g = 4π^2(L/T^2)
Substituting the given values:
g = 4π^2(1.2/330^2)
Calculating this expression will give us the magnitude of the acceleration due to gravity on the planet.
T = 330/100 = 3.30 s. = The Period.
T^2 = 4*pi^2(L/g) = 3.3^2 = 10.89
39.5(1.2/g) = 10.89
47.4/g = 10.89
10.89g = 47.4
g = 4.35 m/s^2
T=330/100=3.3 s
330/100=2pi sqrt (1.2m/g)
rearrange to get :
sqrt G= (pi^2 x sqrt 1.2 )/ (3.3)
(Only square the top part of the fraction to get rid of the sqrt)
G=pi^2 x 1.2 / 3.3
g=3.588 m/s ^2