Question

the mass of a diameter of a planet are twice those of earth. the period of oscillation of pendulum on this planet will be​

Answer 1

Let the mass be M, acceleration due to gravity g, length of pendulum string l, time period be T and universal gravitational constant be G.

Note: All the parameters with subscript ‘2’ are for Earth and those with subscript ‘3’ are for the new planet.

g=GM/r² g=GM/r²

g₂=GM₂/r₂² g₂=GM₂/r₂²

g₃=GM₃/r₃² g₃=GM₃/r₃²

Since M₃=2M₂ M₃=2M₂ andr₃=2r₂ r₃=2r₂

g₃=2GM₂/4r₂² g₃=2GM₂/4r₂²

g₃=GM₂/2r₂² g₃=GM₂/2r₂²

g₃=g₂/2 g₃=g₂/2

NowT=2π√(l/g) T=2π√(l/g)

T₂=2π√(l/g₂) T₂=2π√(l/g₂)

T₃=2π√(l/g₃) T₃=2π√(l/g₃)

T₃=2π√(2l/g₂²) T₃=2π√(2l/g₂²)

T₃=√2.T₂ T₃=√2.T₂

Hence, the time period of a pendulum will become √2 times its original value on the new planet.