Kate (m = 50 kg) is swinging on a tire tied by a (light) rope (L = 3 m) to a tree limb. Her twin Ashley comes along and squeezes into the tire with her. Assume that at all times the centre of mass of the person(s) riding the tire is at the end of the rope.

a) What was the period of oscillation for Kate alone?
b) What is the period of oscillation for the twins together?

I need help please! Do it step by step and then substitute the numbers into that equation

To find the period of oscillation, we need to use the formula for the period of a simple pendulum. The formula is:

T = 2π * sqrt(L/g)

Where:
T is the period of oscillation (in seconds)
L is the length of the pendulum (in meters)
g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)

Let's solve the problem step by step:

a) Period of oscillation for Kate alone:

1. Substitute the given values into the formula:
L = 3 m
g = 9.8 m/s^2

2. Calculate the period:
T = 2π * sqrt(3/9.8)

b) Period of oscillation for the twins together:

1. The center of mass of the twins is still at the end of the rope, so the length of the pendulum remains the same.
L = 3 m

2. Calculate the period using the same formula:
T = 2π * sqrt(3/9.8)

Now, let's substitute the numbers into the equation and calculate the periods for both cases.

For Kate alone:
T = 2π * sqrt(3/9.8)
T ≈ 2π * 0.556
T ≈ 3.49 seconds

For the twins together:
T = 2π * sqrt(3/9.8)
T ≈ 2π * 0.556
T ≈ 3.49 seconds

Both Kate alone and the twins together have the same period of oscillation, which is approximately 3.49 seconds.