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 450 s. What is the magnitude of the acceleration due to gravity on this planet?

use this formula, and memorize it.

Period=2PI sqrt(length/g)

solve for g
Period is 4.5 seconds

10.8

To find the magnitude of the acceleration due to gravity on the distant planet, we can use the formula for the period of a simple pendulum:

T = 2π * √(L / g)

Where:
T is the period in seconds,
L is the length of the pendulum in meters,
g is the acceleration due to gravity in m/s².

In this case, we are given the number of complete oscillations (100) and the time it takes for these oscillations (450 s). We can use these values to first calculate the period T.

T = (time taken for oscillations) / (number of oscillations)

T = 450 s / 100 = 4.5 s

Now we can plug in the known values into the equation for the period and solve for g.

4.5 = 2π * √(1.2 / g)

To solve for g, we need to isolate it on one side of the equation. First, square both sides:

4.5² = (2π)² * (1.2 / g)

20.25 = (4π² / g) * 1.2

Now divide both sides of the equation by 1.2:

20.25 / 1.2 = 4π² / g

16.875 = 4π² / g

Finally, solve for g by taking the reciprocal of both sides of the equation:

g = 4π² / 16.875

g ≈ 9.81 m/s²

Therefore, the magnitude of the acceleration due to gravity on this distant planet is approximately 9.81 m/s².

To find the magnitude of the acceleration due to gravity (g) on the distant planet, we can make use of the formula for the period (T) of a simple pendulum:

T = 2π√(L/g)

Where:
T = Period of the pendulum (in seconds)
L = Length of the pendulum (in meters)
g = Acceleration due to gravity (in m/s²)

In this case, we know the length of the pendulum (L) is 1.2 m and the number of complete oscillations (n) is 100.

First, we can find the period (T) of a single oscillation by dividing the total time (450 s) by the number of complete oscillations (100):

T = 450 s / 100 = 4.5 s

Now, rearranging the formula for T:

T = 2π√(L/g)

We can solve for g:

g = (4π²L) / T²

Plugging in the values:

g = (4π² × 1.2 m) / (4.5 s)²

Calculating:

g = (4 × 3.14159² × 1.2 m) / (4.5 s)²

g ≈ 1.97 m/s²

Therefore, the magnitude of the acceleration due to gravity (g) on this planet is approximately 1.97 m/s².