(a) Your pendulum clock which advances 1.0 s for every complete oscillation of t

Question

(a) Your pendulum clock which advances 1.0 s for every complete oscillation of the pendulum is running perfectly on Earth at a site where the magnitude of the acceleration due to gravity is 9.80 m/s2. You send the clock to a location on the Moon where the magnitude of the acceleration due to gravity is 1.65 m/s2. Does the clock run fast or slow on the Moon?

It runs fast.It runs slow.    Neither, it runs the same as on Earth.

(b) If the clock is started at 12:00 noon, what will it read after a time of 14.2 h has elapsed on Earth? Enter the time to the nearest minute.
:  p.m.

Explanation / Answer

(A) T = 2 pi sqrt(L/g)

T' / T = sqrt(g / g')

T' / 1sce = sqrt(9.8/1.65)

T' = 2.437 sec

now clock will complete single oscillation in 2.437 sec.

so it will run sloe.

(B) oscillations complete in 14.2 h

n = (14.2 x 3600) / (2.437) = 20976.6

so it will clock 20976.6s or 5.83 hrs

so time will be 5 : (0.83 x 60)

= 5 : 50

( 5 hr 50 sec)

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