JEE Advanced Model Paper 11 with solutions for online practice

Due to the extension the additional tension in each spring will be f.
The restoring force due to these two springs
Frestoring = 2 (F0 + f) sin θ
⇒ Frestoring = 2 F0 sin θ ( ∵ f ≪ F )
⇒ Frestoring = 2 F0 θ {Assume θ to be small}
⇒ Frestoring = 2 F0 (

y

t

) {∵ sin θ =

y

l

}

Net restoring force due to all four springs is
Fnet = 4 F0 (

y

l

) {directed towards the mean position}