Oscillations Part 1

Given that, a bob of mass m suspended by a thread of length l undergoes SHM with time period,
T1=T=2π√

l

g

...(i)
In liquid, effective gravity,
geff=g(

σ−ρ

σ

)
where, ρ = density of liquid
and σ = density of body (bob).
Given, ρ=

σ

4

⇒σ=4ρ
Substituting this value, we get
geff=g(

4ρ−ρ

4ρ

)=

3g

4

Now, the new time period of pendulum in liquid is
T′=2π√

l′

geff

...(ii)
Here,l′=l+

l

3

=

4I

3

Substituting the values in Eq. (ii), we get
T′=2π√

4l

3

×

4

3g

=(2π√

l

g

)

4

3

⇒T′=

4T

3

[from Eq. (i)]