JEE Advanced Model Paper 2 with solutions for online practice

ln the figure, F = -mg sin(θ1+θ2) for small amplitude sin (θ1+θ2) = θ1+θ2

∴ F = -mg (θ1+θ2)
F = - mg (sin θ1 + sin θ2) = -mg (

x

L

+

x

Re

) = -mg (

1

L

+

1

Re

) x
∴ a = - g (

1

L

+

1

Re

) x
Comparing with a = - ω2 x ,
ω = √g(

1

L

+

1

Re

)
∴ T = 2π √

1

g( 1 L + 1 Re )

When L = Re i.e = L = 6400 × 103 m
We get T = 59.8 min
When L → ∞ ,

Re

L

= 0
Therefore , T = 2π √

Re

g

= 84.6 min