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Question: The bob of a simple pendulum performs S.H.M. with period ‘T’ in air and with period ‘\[{T_1}\]’ in w...

The bob of a simple pendulum performs S.H.M. with period ‘T’ in air and with period ‘T1{T_1}’ in water. Relation between ‘T’ and ‘T1{T_1}’ is (neglect the friction due to water, density of the material of the bob is 98×103kg/m3\dfrac{9}{8} \times {10^3}kg/{m^3} , density of water =1g/cc1g/cc)
(A) T1=3T{T_1} = 3T
(B) T1=2T{T_1} = 2T
(C) T1=T{T_1} = T
(D) T1=T2{T_1} = \dfrac{T}{2}

Explanation

Solution

Hint Consider two cases of time periods. Calculate the effective g and find its relation with g in air. Since given data is in terms of density the mass is converted in terms of density as well. Then substitute the given data to obtain a relation between the two time periods.

Complete step-by-step solution

T and T1{T_1}are the time period of the simple pendulum in air and water respectively.

We know that the time period of the pendulum is,
T=2πlgT = 2\pi \sqrt {\dfrac{l}{g}}
The length of the pendulum l remains the same in both mediums but g is not the same so we need to consider an effective value
ge{g_e} forT1{T_1}
ge{g_e} is given by the difference between the weight of the bob and the buoyancy force when in water.
ge=mgfbm{g_e} = \dfrac{{mg - {f_b}}}{m}
Here,
m is the mass of the bob
g is the acceleration due to gravity
Fb{F_b} is the buoyancy force= mass of water displaced × g \times {\text{ }}g

Mass is written in terms of density ρ\rho using the formula,
ρ=mV\rho = \dfrac{m}{V}
ge=ρVgρ1VgρV=g(ρρ1)ρ{g_e} = \dfrac{{\rho Vg - {\rho _1}Vg}}{{\rho V}} = \dfrac{{g\left( {\rho - {\rho _1}} \right)}}{\rho }
Where,
ρ\rho is the density of the material of the bob 98×103kg/m3\dfrac{9}{8} \times {10^3}kg/{m^3}

ρ1{\rho _1} is the density of water 1g/cc1g/cc
On substituting the values,
ge=(98×1031×103)g98×103 ge=g9  {g_e} = \dfrac{{\left( {\dfrac{9}{8} \times {{10}^3} - 1 \times {{10}^3}} \right)g}}{{\dfrac{9}{8} \times {{10}^3}}} \\\ {g_e} = \dfrac{g}{9} \\\

Since,
T1=2πlge{T_1} = 2\pi \sqrt {\dfrac{l}{{{g_e}}}}
Now, substitute the value of ge{g_e} in the above equation, we get,
T1=2πl9g=3(2πlg) T1=3T  {T_1} = 2\pi \sqrt {\dfrac{{l9}}{g}} = 3\left( {2\pi \sqrt {\dfrac{l}{g}} } \right) \\\ \therefore {T_1} = 3T \\\

Hence, the correct option is A.

Note The pendulum in water displaces some volume so it has to be considered to calculate the time period. Time period is the time taken to complete one oscillation about the mean position. It depends on the medium in which it is oscillating.