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

Answer:

$T_{m }$ = 4.86 s

$T_{e}$ = 1.98 s

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 $\sqrt{\frac{l}{g} }$

Solution:

For moon

Putting the givens,

T = 2(3.14) $\sqrt{\frac{1}{1.67} }$

T = 6.3 $\sqrt{0.6}$

T = 6.3 × 0.77

T = 4.86 sec

For Earth,

Putting the givens

T = 2π $\sqrt{\frac{1}{10} }$

T = 2(3.14) $\sqrt{0.1}$

T = 6.3 × 0.32

T = 1.98 sec