Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 51.0 cm

. The explorer finds that the pendulum completes 110 full swing cycles in a time of 133 s
.
Part A
What is the magnitude of the gravitational acceleration on this planet?
Express your answer in meters per second per second.

Answer 1

Answer: 7.68 m/s²

Explanation: The period of a simple pendulum is given by the formula:

T = 2π√(L/g)

where T is the period, L is the length of the pendulum, and g is the gravitational acceleration.

In this problem, the pendulum completes 110 full swing cycles in a time of 133 s, which means that the period of the pendulum is:

T = 133 s / 110 = 1.209 s

Substituting this into the formula above and solving for g, we get:

g = (4π²L) / T² = (4π²)(0.51 m) / (1.209 s)² ≈ 7.68 m/s²

Therefore, the magnitude of the gravitational acceleration on this planet is approximately 7.68 m/s².

Answer 2

The period of a simple pendulum is given by the formula:

T = 2π√(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.

We can rearrange this formula to solve for g:

g = (4π²L)/T²

Plugging in the given values:

L = 51.0 cm = 0.510 m

T = 133 s / 110 = 1.209 s

(Note that we divide the total time by the number of cycles to get the time for one cycle.)

So,

g = (4π² × 0.510 m) / (1.209 s)²

= 9.57 m/s²

Therefore, the magnitude of the gravitational acceleration on this planet is approximately 9.57 m/s².

*IG:whis.sama_ent*