Question

A simple pendulum with a metal bob has a period T. What will be the period of the same pendulum if it is made to oscillate in a non-viscous liquid of density one-tenth of the metal of the bob?

Answer 1

Hint: Archimedes principle states that any body partially or fully submerged in a liquid or gas at the state of rest is acted upon by an upward buoyant force the magnitude of which is equal to the fluid displaced by the body.
The time period of a pendulum is given by $T = 2\pi \sqrt {\dfrac{l}{g}} $ where T is the time period, l is the length and g is the acceleration due to gravity.
If the pendulum is immersed in a liquid, then it will also experience a buoyant force and it will change the value of g acting on it.
The formula then becomes $T = 2\pi \sqrt {\dfrac{l}{{{g_{eff}}}}} $ where ${g_{eff}}$ is the effective value of acceleration due to gravity acting on the metal bob.

Complete step-by-step answer:
The metal bob is immersed in a non-viscous liquid of density one-tenth of the metal of the bob. Hence,
$m{g_{eff}} = mg - \,{F_b}$ where ${F_b}$ denotes the buoyant force.
\[ \Rightarrow {g_{eff}} = g - \dfrac{{\,{F_b}}}{m}\]
The buoyant force is given as ${F_b} = V\rho g$ where V is the volume of the object, $\rho $ is the density of the object and g is the acceleration due to gravity.
So, the equation becomes \[{g_{eff}} = g - \dfrac{{\,V{\rho _{liquid}}g}}{m}\]
Now $m = V\rho $ and it is given that ${\rho _{liquid}} = \dfrac{\rho }{{10}}$ .
So, the equation reduces to
\[{g_{eff}} = g - \dfrac{{\,V \times \dfrac{\rho }{{10}} \times g}}{{V\rho }}\]
Solving this further,
\[{g_{eff}} = g - \dfrac{g}{{10}}\]
\[{g_{eff}} = \dfrac{{9g}}{{10}}\]
Now we know that the time period of a pendulum is given by $T = 2\pi \sqrt {\dfrac{l}{g}} $
When the bob is immersed in the non-viscous liquid the effective time period is $T' = 2\pi \sqrt {\dfrac{l}{{{g_{eff}}}}} $
Substituting the value in this equation we get,
$T' = 2\pi \sqrt {\dfrac{l}{{\dfrac{{9g}}{{10}}}}} $
Further simplifying,
$T' = \dfrac{{\sqrt {10} }}{3}2\pi \sqrt {\dfrac{l}{g}} $
$ \Rightarrow T' = \dfrac{{\sqrt {10} }}{3}T$

Note: Very often weight is misunderstood as mass of the object. These are two different terms and must be noted carefully. Mass is the inherent property of an object which tells how much matter it contains while weight is the normal reaction acting on the body. Mass is constant while the weight may vary.