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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

Respuesta :

hamzaahmeds

Answer:

c. More than 2 s

Explanation:

First, we will find the length of the pendulum:

[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\2\ s = 2\pi \sqrt{\frac{l}{9.81\ m/s^2}}\\\\4\ s^2 = 4\pi^2 (\frac{l}{9.81\ m/s^2})\\\\l = \frac{(4\ s^2)(9.81\ m/s^2)}{4\pi^2} \\\\l = 0.99\ m[/tex]

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

[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\T = 2\pi \sqrt{\frac{0.99\ m}{1.625\ m/s^2}}\\\\[/tex]

T = 4.9 s

Therefore, the correct option is:

c. More than 2 s

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