Question

calculate the time period of simply pendulum of length 1.44m on the surface of moon. The acceleration due to gravity on the surface of moon is 1/6 the acceleration due to gravity on earth (g=9.8ms-2)​

Answer 1

Answer:

Acceleration due to gravity on the surface of moon, g

′

=1.7ms

−2

Acceleration due to gravity on the surface of earth, g=9.8ms

−2

Time period of a simple pendulum on earth, T=3.5s

T=2π

g

l

where,

l is the length of the pendulum

∴l=

(2π)

2

T

2

×2

=

4×(3.14)

2

(3.5)

2

×9.8m

The length of pendulum remains constant

On moon's surface, time period, T

′

=2π

g

′

l

=2π

1.7

4×(3.14)

2

(3.5)

2

×9.8

=8.4 s

Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.