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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 .

Respuesta :

Answer:

g / 16

Explanation:

T = 2π [tex]\sqrt{\frac{l}{g} }[/tex]

angular frequency ω = 2π /T

= [tex]\sqrt{\frac{g}{l} }[/tex]

ω₁ /ω₂ = [tex]\sqrt{\frac{g_1}{g_2} }[/tex]

Putting the values

ω₁ = ω ,     ω₂ = ω / 4

ω₁ /ω₂ = 4

4 =  [tex]\sqrt{\frac{g}{g_2} }[/tex]

g₂ = g / 16

option d is correct.

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