Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 49.0 cm. The explorer finds that the pendulum completes 109 full swing cycles in a time of 144 s.What is the value of the acceleration of gravity on this planet?

Answer 1

The acceleration due to gravity on the planet is approximately 45.6 m/s².

Step-by-step explanation:

The period of a simple pendulum can be calculated using the formula:

Period = 2 * π * sqrt(length / acceleration due to gravity)

Given that the pendulum completes 109 full swing cycles in a time of 144 s, we can find the period by dividing the total time by the number of cycles: Period = time / number of cycles = 144 s / 109 = 1.3211 s.

Now, using the formula for the period, we can solve for the acceleration due to gravity:

1.3211 s = 2 * π * sqrt(0.49 m / acceleration due to gravity)

Simplifying the equation, we find that the acceleration due to gravity on this planet is approximately 45.6 m/s².

Answer 2

The frequency refers to the number of cycles per time lapse, while the period is the inverse of the frequency, therefore we would have this expression,

`f= (N)/(t)`

Here,

N = Number of Cycles

t = Time

While at the same time Period is the inverse of the frequency, then

`T = (1)/(f)`

Replacing,

`T = (1)/(N/t)`

`T = (t)/(N)`

Replacing we have that

`T = (144s)/(109)`

`T = 1.321s`

Gravity is one of the components that define the Period of a pendulum as well as its length, mathematically this expression is given as,

`T = 2\pi \sqrt{(l)/(g)}`

`g = (4\pi^2l)/(T^2)`

Replacing,

`g = (4\pi^2 (49*10^(-2)m))/((1.321s^2)^2)`

`g = 11.0853m/s^2`

Therefore the acceleration due to gravity of the planet is
`11.08m/s^2`