(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