The correct option is
C 4π3√pbapwg+pbpw−pb√4pwa3pbgIn case of equilibrium,
B=mg
ρ(y)A)g=ρbaAg
⇒y0=ρbaρw
If we displace the body in downdwards direction, then
⇒Frestoring=−(B−mg)
⇒Frestoring=−(ρw(y0+y)Ag−ρbAag)
⇒Frestoring=−ρwy0Ag−ρwyAg+ρbAag
Substituting value of
y0 we get,
Frestoring=−ρbAag−ρwyAg+ρbAag
∴Frestoring=−(ρwAg)y
Since restoring force is proportional to displacement from mean position, hence body is performing SHM motion.
So we can write,
Frestoring=−mω2y=−(ρwAg)y
⇒ρbAaω2y=ρwAgy
⇒ω=√ρwgρba
Now since
y0=2a3 ,
This situation can be better understood from the diagram below.
So amplitude can be calculated as
A=a2+a6=2a3
But by the time body comes down
a3, it will be completely submerged, and from there on buoyancy force does not increase more. So the motion ceases to be SHM.
So phasor of the motion of body will be,
cosϕ=yA=a32a3=12
⇒ϕ=π3
So body is performing SHM only for the phase of
2π−2ϕ=4π3.
Hence time duration when body is performing SHM will be
T1=4π3√ρbaρwg
Now for time taken for uniformly accelerated motion:
a=B−mgm
a=ρw−ρbρbg
Velocity when body has just reached its completely submerged position,
v=ω√A2−y2
v=√ρwgρba√4a29−a29
v=√ρwga3ρb
So time taken during uniformly accelerated motion,
T2=2va=2√ρwa3ρbgρbρw−ρb
So total time
=T1+T2=4π3√pbapwg+pbpw−pb√4pwa3pbg