Question

You have a simple pendulum that oscillates with a period of 2 s as you stand on the surface of Earth. Your friend, an astronaut standing on the surface of the Moon, has a pendulum of the same length. What would be the period of oscillation of your friend’s pendulum?

a. Less than 2 s
b. The answer depends on whether the amplitudes are the same
c. More than 2 s
d. Exactly 2 s

Answer 1

c. More than 2 s

Step-by-step explanation:

First, we will find the length of the pendulum:

`T = 2\pi \sqrt{(l)/(g)}\\\\2\ s = 2\pi \sqrt{(l)/(9.81\ m/s^2)}\\\\4\ s^2 = 4\pi^2 ((l)/(9.81\ m/s^2))\\\\l = ((4\ s^2)(9.81\ m/s^2))/(4\pi^2) \\\\l = 0.99\ m`

Now, the value of g becomes 1.625 m/s² on the surface of the moon. So the time period will be:

`T = 2\pi \sqrt{(l)/(g)}\\\\T = 2\pi \sqrt{(0.99\ m)/(1.625\ m/s^2)}\\\\`

T = 4.9 s

Therefore, the correct option is:

c. More than 2 s