Question

The period of oscillation for a pendulum on Earth is 9 seconds. If the given pendulum oscillates with a period of 22.1 secondson the surface of the Moon, what is the acceleration due to gravity on the Moon's surface

Answer 1

1.625 m/s²

Step-by-step explanation:

The acceleration due to gravity gotten from the period of the simple pendulum using the formula:

`T=2\pi \sqrt{(L)/(g) }\\ \\where\ T=period,L=length\ of \ pendulum,g=acceleration\ due\ to\ gravity.\\\\For \ earth, g=9.8\ m/s^2.Given\ that\ period(T)=9\ seconds\ we\ can\ find\ L:\\\\T=2\pi \sqrt{(L)/(g) }\\\\squaring\ both\ sides:\\\\T^2=4\pi^2((L)/(g) )\\\\L=(T^2g)/(4\pi^2) \\\\Substituting\ gives:\\\\L=(9^2*9.8)/(4\pi^2) =20.1\ m\\\\The\ same\ pendulum\ is\ used \ in \ moon\ and \ gives\ a\ period\ of\ 22.1\ seconds. \\`

`We\ can\ get\ the\ acceleration\ due\ to\ gravity\ on\ the\ moon\ using:\\\\T=2\pi \sqrt{ (L)/(g)}\\ \\T^2=4\pi^2 ((L)/(g) )\\\\g=(4\pi^2 L)/(T^2)\\ \\g=(4\pi^2 *20.1)/(22.1^2)\\\\g=1.625\ m/s^2`