Question

A simple pendulum has a time period T. The

pendulum is cornpletely immersed in a


non viscous liquid ,whose density is 1/20th of that

of the material of the bob. Then the time period

of the pendulum immersed in the liquid is:
​

Answer 1

So the simple pendulum has a time period,

• $\sf{T=2\pi\sqrt{\dfrac{m}{k}}}$

Recall $\sf{kl=mg.}$

Now the pendulum is completely immersed in a  non viscous liquid, whose density is $\sf{\left(\dfrac{1}{20}\right)^{th}}$ of that  of the material of the bob.

• $\sf{\rho_L=\dfrac{\rho_B}{20}}$

The fig. shows the FBD of the bob at mean position.

• $\sf{mg}$ is weight of bob.

• $\sf{k'l}$ is restoring force produced in the string of length $\sf{l.}$

• $\sf{U}$ is upthrust.

We get,

$\sf{\longrightarrow mg=k'l+U}$

$\sf{\longrightarrow k'l=mg-U}$

Let $\sf{V}$ be volume of bob. So $\sf{m=\rho_BV.}$

The upthrust $\sf{U}$ equals weight of water displaced by the bob. So,

$\sf{\longrightarrow k'l=\rho_BVg-\rho_LVg}$

$\sf{\longrightarrow k'l=\left(\rho_B-\rho_L\right)Vg}$

$\sf{\longrightarrow k'l=\left(\rho_B-\dfrac{\rho_B}{20}\right)Vg}$

$\sf{\longrightarrow k'l=\dfrac{19}{20}\,\rho_BVg}$

$\sf{\longrightarrow k'l=\dfrac{19}{20}\,mg}$

$\sf{\longrightarrow k'l=\dfrac{19}{20}\,kl}$

$\sf{\longrightarrow k'=\dfrac{19}{20}\,k}$

Hence the new time period,

$\sf{\longrightarrow T'=2\pi\sqrt{\dfrac{m}{k'}}}$

$\sf{\longrightarrow T'=2\pi\sqrt{\dfrac{20}{19}\cdot\dfrac{m}{k}}}$

$\sf{\longrightarrow\underline{\underline{T'=T\sqrt{\dfrac{20}{19}}}}}$

Answer 2

So the simple pendulum has a time period,

\sf{T=2\pi\sqrt{\dfrac{m}{k}}}T=2π

k

m

Recall \sf{kl=mg.}kl=mg.

Now the pendulum is completely immersed in a non viscous liquid, whose density is \sf{\left(\dfrac{1}{20}\right)^{th}}(

20

1

)

th

of that of the material of the bob.

\sf{\rho_L=\dfrac{\rho_B}{20}}ρ

L

=

20

ρ

B

The fig. shows the FBD of the bob at mean position.

\sf{mg}mg is weight of bob.

\sf{k'l}k

′

l is restoring force produced in the string of length \sf{l.}l.

\sf{U}U is upthrust.

We get,

\sf{\longrightarrow mg=k'l+U}⟶mg=k

′

l+U

\sf{\longrightarrow k'l=mg-U}⟶k

′

l=mg−U

Let \sf{V}V be volume of bob. So \sf{m=\rho_BV.}m=ρ

B

V.

The upthrust \sf{U}U equals weight of water displaced by the bob. So,

\sf{\longrightarrow k'l=\rho_BVg-\rho_LVg}⟶k

′

l=ρ

B

Vg−ρ

L

Vg

\sf{\longrightarrow k'l=\left(\rho_B-\rho_L\right)Vg}⟶k

′

l=(ρ

B

−ρ

L

)Vg

\sf{\longrightarrow k'l=\left(\rho_B-\dfrac{\rho_B}{20}\right)Vg}⟶k

′

l=(ρ

B

−

20

ρ

B

)Vg

\sf{\longrightarrow k'l=\dfrac{19}{20}\,\rho_BVg}⟶k

′

l=

20

19

ρ

B

Vg

\sf{\longrightarrow k'l=\dfrac{19}{20}\,mg}⟶k

′

l=

20

19

mg

\sf{\longrightarrow k'l=\dfrac{19}{20}\,kl}⟶k

′

l=

20

19

kl

\sf{\longrightarrow k'=\dfrac{19}{20}\,k}⟶k

′

=

20

19

k

Hence the new time period,

\sf{\longrightarrow T'=2\pi\sqrt{\dfrac{m}{k'}}}⟶T

′

=2π

k

′

m

\sf{\longrightarrow T'=2\pi\sqrt{\dfrac{20}{19}\cdot\dfrac{m}{k}}}⟶T

′

=2π

19

20

⋅

k

m

\sf{\longrightarrow\underline{\underline{T'=T\sqrt{\dfrac{20}{19}}}}}⟶

T

′

=T

19

20