Question

find the period of a simple pendulum of 1m length placed on earth and on moon g on moon =1.67m/s² g on earth=10m/s²

Answer 1

`T_(m )` = 4.86 s

`T_(e)` = 1.98 s

Step-by-step explanation:

Given:

Length = l = 1 m

Acceleration due to gravity of moon =
`g_(m)` = 1.67 m/s²

Acceleration due to gravity of Earth =
`g_(e)` = 10 m/s²

Required:

Time period = T = ?

Formula:

T = 2π
`\sqrt{(l)/(g) }`

Solution:

For moon

Putting the givens,

T = 2(3.14)
`\sqrt{(1)/(1.67) }`

T = 6.3
`√(0.6)`

T = 6.3 × 0.77

T = 4.86 sec

For Earth,

Putting the givens

T = 2π
`\sqrt{(1)/(10) }`

T = 2(3.14)
`√(0.1)`

T = 6.3 × 0.32

T = 1.98 sec