Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of

length 1.51 m and finds that it makes 291 complete oscillations in 697 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/s2

thanks!!!!

Well, just like a comedian landing on a new planet, let's do some calculations to find the punchline!

We can start by using the formula for the period (T) of a pendulum:

T = 2π√(L/g)

Where L is the length of the pendulum and g is the acceleration due to gravity.

Now, we know that the astronaut observed 291 complete oscillations in 697 seconds. This means the period of one oscillation (T) is:
T = 697 s / 291 = 2.395 s

Plugging this value into the pendulum formula:

2.395 s = 2π√(1.51 m / g)

To find g, we can rearrange the equation:

g = (2π)^2 * (1.51 m) / (2.395 s)^2

Now, let's crunch the numbers:

g ≈ 9.8 m/s²

Ah, good old gravity! The gravitational acceleration on the surface of this newly discovered planet is approximately 9.8 m/s², just like our good old Earth. So, if you ever find yourself on that planet, remember to stay grounded!

To find the gravitational acceleration on the surface of the planet, we can use the formula for the period (T) of a pendulum:

T = 2 * pi * sqrt(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 know the period (T), the length (l), and we need to find the acceleration due to gravity (g).

From the problem, we are given:
Length of the pendulum (l) = 1.51 m
Number of oscillations in the given time (N) = 291
Time (T) = 697 s

Using the formula for the period of a pendulum, we can rearrange it to solve for g:

g = 4 * pi^2 * l / T^2

Plugging in the given values:

g = 4 * pi^2 * 1.51 m / (697 s)^2

Calculating this, we find:

g ≈ 6.834 m/s^2

Therefore, the gravitational acceleration on the surface of the newly discovered planet is approximately 6.834 m/s^2.

To find the gravitational acceleration on the surface of the newly discovered planet, we can use the formula for the period of a simple pendulum:

T = 2π * √(L/g)

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

In this case, we are given the length of the pendulum (L = 1.51 m) and the number of oscillations in a given time (291 oscillations in 697 s), which can be used to calculate the period of the pendulum.

First, we calculate the time taken for one oscillation:

Time per oscillation = Total Time / Number of Oscillations
Time per oscillation = 697 s / 291 oscillations

Next, we calculate the period of the pendulum:

T = Time per oscillation / Number of oscillations
T = (697 s / 291 oscillations) / 291 oscillations

Now that we have the period, we can rearrange the formula to solve for the gravitational acceleration:

g = (4π² * L) / T²

Plugging in the given values, we can calculate the gravitational acceleration:

g = (4π² * 1.51 m) / (T²)

Remember to convert the period to seconds:

T = (Time per oscillation / Number of oscillations) in seconds

Finally, calculate the gravitational acceleration:

g = (4π² * 1.51 m) / (T in seconds)²

By solving this equation, you will find the gravitational acceleration on the surface of the newly discovered planet in units of m/s².