JEE Advanced Model Paper 1 with solutions for online practice

(i) F = -[k(x + x0) - mg]
or F = -kx (∵ at equilibrium mg = kx0)
a = −

k

m

x
ω = √

k

m

T1 = 2π √

m

k

(ii) In this case, if the mass m moves down a distance x from its equilibriumposition, then pulley will move down by

x

2

So, the extra force in the spring will be

kx

2

F = −

k

4

x
a = −

kx

4m

⇒ ω = √

k

4m

∴ T2 = 2π √

4m

k

iii) In this case, if the mass 'm' moves down a distance 'x' from it's equilibriumposition, then pulley will also move by x and so the spring will stretch by 2x.Restoring force on the block is 2x

a =

−4x

m

ω = √

4kx

m

T = 2π √

m

4k