Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cm. The explorer finds that the pendulum completes 98.0 full swing cycles in a time of 135s.
Required:
What is the value of g on this planet?

Answer 1

Answer:

g = 11.2 m/s²

Explanation:

First, we will calculate the time period of the pendulum:

$T = \frac{t}{n}$

where,

T = Time period = ?

t = time taken = 135 s

n = no. of swings in given time = 98

Therefore,

$T = \frac{135\ s}{98}$

T = 1.38 s

Now, we utilize the second formula for the time period of the simple pendulum, given as follows:

$T = 2\pi \sqrt{\frac{l}{g}}$

where,

l = length of pendulum = 54 cm = 0.54 m

g = acceleration due to gravity on the planet = ?

Therefore,

$(1.38\ s)^2 = 4\pi^2(\frac{0.54\ m}{g} )\\\\g = \frac{4\pi^2(0.54\ m)}{(1.38\ s)^2}$

g = 11.2 m/s²