Question

A pendulum clock that works perfectly on Earth is taken to the Moon. Assume that the free-fall acceleration on the Moon is 1.63 m/s².

(a) Does it run fast or slow there?

fastslow    Neither, it runs the same as on Earth.

(b) If the clock is started at 12:00 midnight, what will it read after 17.6 h? (Enter the time to the nearest minute.)
:  AM

Answer 1

(a) Slows down

Period is inversely proportional to g. g of moon is less than g of Earth. So, as g is less for moon, so period is longer and thus it slows down.

(b) Are those earth times? 17.6 hours is 1056 minutes

Assuming the pendulum is set for a period of 1 second on earth

1 = 2π√(L/g) = 2π√(L/9.8)

L/9.8 = 1/4π²

L = 9.8/4π²

now g changes to 1.63

T = 2π√(L/g) = 2π√(9.8/4π²1.63) = √(9.8/1.63) = 2.45 seconds or a factor of 2.45

that is per tick.

1056 x 2.45 = 2587.2 min, which is 43.12 hours or 43 hours 7 min

Pendulum period in seconds

T ≈ 2π√(L/g)

or, rearranging:

g ≈ 4π²L/T²

L ≈ T²g/4π²

L is length of pendulum in meters

g is gravitational acceleration = 9.8 m/s²