Question

Under the conditions of L=0.5m; m=0.5kg; Theta=15 degrees, the period of oscillation on the moon is approximately equal to:

a) 2.24 seconds
b) 1.58 seconds
c) 1.12 seconds
d) 0.79 seconds

Answer 1

The period of oscillation on the moon under the given conditions is approximately equal to 1.58 seconds.

Step-by-step explanation:

To find the period of oscillation on the moon under the given conditions, we can use the formula for the period of a simple pendulum:

T = 2π√(L/g)

where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. In this case, L = 0.5m, m = 0.5kg, and Θ = 15 degrees. The acceleration due to gravity on the moon is 1.63 m/s2.

Plugging in the values, we have:

T = 2π√(0.5/1.63) = 1.58 seconds

Therefore, the period of oscillation on the moon is approximately equal to 1.58 seconds.