Having landed on a newly discovered planet, an astronaut sets up a simple pedulum of length 1.29 m and finds that it makes 567 oscillations in 1020 s. The amplitude of the oscillations is very small compared to the pendulum's length.

What is the gravitational acceleration on the surface of this planet? (Answer in units of m/s squared)

This is a duplicate post. I answered it yesterday.

Use the formula for the period:

P = 2 pi sqrt (L/g') and solve for g'

The period is 1020/567 s. Convert to decimal.

To find the gravitational acceleration on the surface of the planet, you can use the formula for the period of a simple pendulum. The formula is given by:

T = 2π √(L/g)

Where:
T = Period of the pendulum
L = Length of the pendulum
g = Gravitational acceleration

In this case, we know the length of the pendulum (L = 1.29 m) and the total time for 567 oscillations (T = 1020 s). The total time is equal to the number of oscillations (567) multiplied by the period of one oscillation (T/567).

So, first, let's calculate the period of one oscillation:

Period of one oscillation = T/567 = 1020 s / 567 = 1.8 s

Now, we can substitute the values into the formula for the period of a pendulum and solve for g:

1.8 s = 2π √(1.29 m / g)

To isolate the variable g, we need to square both sides of the equation:

(1.8 s)² = (2π)² * (1.29 m / g)

3.24 s² = 4π² * (1.29 m / g)

Now, isolate g:

g = 4π² * (1.29 m) / (3.24 s²)

Calculating the value:

g = (4 * 3.14159² * 1.29 m) / (3.24 s²)

g ≈ 15.43 m/s²

Therefore, the gravitational acceleration on the surface of this planet is approximately 15.43 m/s².