Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.29 m and finds that it makes 407 complete oscillations in 698 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
Use the small-amplitude formula:
Period = 2*pi*sqrt(L/g)
and solve for g.
L = 1.29 m
Period P = 698/407 = 1.715 seconds
g/L = 4*pi^2/P^2 = 13.42 s^-2
g = 17.31 m/s^2
Well, well, well, an astronaut and a pendulum walk into a newly discovered planet... talk about an out-of-this-world party! Now, to calculate the gravitational acceleration on this funky planet, we need to use a little formula called the period of a pendulum.
T = 2π√(L/g)
Where T is the period of the pendulum, L is the length of the pendulum, and g is the gravitational acceleration.
In this case, we know the length of the pendulum (1.29 m) and the number of oscillations in a certain time (407 oscillations in 698 s).
By dividing the total time (698 s) by the number of oscillations (407), we get the time for one oscillation (T). Then, we can simply plug the values into the equation and solve for g.
But hey, I don't want to leave you hanging! Let me crunch the numbers for you. After some quick calculations, the gravitational acceleration on the surface of this planet is approximately equal to 6.91 m/s².
So, on this planet, you might feel a little lighter than usual, but don't worry, that just means you can eat more space donuts! Enjoy your adventure!
To find the gravitational acceleration on the surface of this planet, we can use the formula 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
g is the gravitational acceleration
In this case, we know the length of the pendulum is 1.29 m and the number of oscillations in a given time is 407 in 698 s. The period can be calculated by dividing the total time by the number of oscillations:
T = 698 s / 407
Now we can substitute the known values into the formula:
698 s / 407 = 1.713 s (approx)
Now, we can rearrange the formula to solve for g:
g = (4π²L) / T²
Substituting the values:
g = (4π² * 1.29 m) / (1.713 s)²
Calculating this:
g ≈ (4 * 3.1416² * 1.29) / (1.713²)
g ≈ 14.75 m/s²
Therefore, the gravitational acceleration on the surface of this planet is approximately 14.75 m/s².
To find the gravitational acceleration on the surface of the planet, we can use the formula for the period of a simple pendulum, which is given by:
T = 2π * √(L/g)
where T is the period, L is the length of the pendulum, and g is the gravitational acceleration.
In this case, the period, T, can be calculated by dividing the total time, 698 s, by the number of oscillations completed, 407:
T = 698 s / 407 = 1.715 s
Now, substituting the known values into the formula, we can solve for g:
1.715 s = 2π * √(1.29 m / g)
To isolate g, we need to rearrange the equation:
√(1.29 m / g) = 1.715 s / (2π)
Squaring both sides, we get:
1.29 m / g = (1.715 s / (2π))^2
Simplifying:
g = 1.29 m / [(1.715 s / (2π))^2]
Calculating the right-hand side of the equation gives us:
g ≈ 10.0 m/s²
Therefore, the gravitational acceleration on the surface of this planet is approximately 10.0 m/s².