Question

Compare the time periods of a simple pendulum at places where g = 9.8 ms⁻² and 4.36 ms⁻² respectively.

Answer 1

g₁ = 9.8 ms⁻²

g₂ = 4.36 ms⁻²

Now, time period of pendulum is, `"T" = 2root(pi)(l/g)`

`"T"_1 = 2root(pi)(l/(g_1))`

`"T"_1 = 2root(pi)(l/9.8)`

= 2π × 0.3194 × 1 = 2.0061

Similarly,

`"T"_2 = 2root(pi)(l/(g_2))`

`"T"_2 = 2root(pi)(l/(4.36))`

= 2π × 0.4789 × 1 = 3.0071

T₁ : T₂ = 2.0061 : 3.0071

= 0.667 : 1