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

Answer:

g / 16

Explanation:

T = 2π $\sqrt{\frac{l}{g} }$

angular frequency ω = 2π /T

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

ω₁ /ω₂ = $\sqrt{\frac{g_1}{g_2} }$

Putting the values

ω₁ = ω ,     ω₂ = ω / 4

ω₁ /ω₂ = 4

4 =  $\sqrt{\frac{g}{g_2} }$

g₂ = g / 16

option d is correct.