Question

A pendulum on earth swings with angular frequency ω. On an unknown planet, it swings with angular frequency ω/ 4. The acceleration due to gravity on this planet is A pendulum on earth swings with angular frequency . On an unknown planet, it swings with angular frequency 4. The acceleration due to gravity on this planet is a. 16 g. b. 4 g.c. g/4. d. g/16 .

Answer 1

g / 16

Step-by-step explanation:

T = 2π
`\sqrt{(l)/(g) }`

angular frequency ω = 2π /T

=
`\sqrt{(g)/(l) }`

ω₁ /ω₂ =
`\sqrt{(g_1)/(g_2) }`

Putting the values

ω₁ = ω , ω₂ = ω / 4

ω₁ /ω₂ = 4

4 =
`\sqrt{(g)/(g_2) }`

g₂ = g / 16

option d is correct.