Question

The amplitude of a simple pendulum that is oscillating in air with a small spherical bob is found to decrease from 10cm to 8cm in the duration of 40s. Let us Assume that Stokes law is valid and the ratio of the coefficient of viscosity of air to that of carbon dioxide is given by 1 : 3. The duration of time during which the amplitude of this particular pendulum will be reducing from 10cm to 5 cm in carbon dioxide will be close to (ln 5 = 1.601, ln 2 = 0.693)

Answer 1

Hint: You may recall the relationship between the amplitude of damped oscillation with respect to time. With the help of Stoke’s law, you could also find the value of damping constant and then you could write the equation for amplitude in both of these cases. Thereby, you will be able to find the required time taken.

Formula used:
$A={{A}_{0}}{{e}^{\dfrac{-bt}{2m}}}$
F=-bv
$F=-6\pi \eta rv$

Complete answer:
Due to air resistance, the bob of the pendulum undergoes damped oscillations. As a result, the amplitude of the oscillations decreases with time.
The amplitude of damped oscillations at time t is given as $A={{A}_{0}}{{e}^{\dfrac{-bt}{2m}}}$ …. (i).
Here, ${{A}_{0}}$ is the amplitude at time t=0 (i.e. the amplitude at which the oscillations begin), b is damping constant and m is the mass of the bob.
The air resistance is due to the viscosity of air. As a result, the air exerts force called damping force on the bob which opposes its motion.
The damping force is given as F = -bv…… (ii),
Where, b is the damping constant and v is the velocity of the bob.
From the Stokes law, we know that due to the viscosity, the bob will experience a drag force which is given as $F=-6\pi \eta rv$….. (iii), where r is the radius of the bob, $\eta $ is the coefficient of viscosity of the medium
In this case, the drag force acts as the damping force.
Therefore, on equating the equations (ii) and (iii), we get
$b=6\pi \eta r$.
Substitute the value of b in equation (i).
$\Rightarrow A={{A}_{0}}{{e}^{\dfrac{-6\pi \eta rt}{2m}}}$ …. (iv)
When the pendulum is in air, we can write its amplitude as
$\Rightarrow A={{A}_{0}}{{e}^{\dfrac{-6\pi {{\eta }_{1}}rt}{2m}}}$.
It is given that the amplitude changes from 10cm to 8cm in the 40s.
$\Rightarrow 8=10{{e}^{\dfrac{-6\pi {{\eta }_{1}}r(40)}{2m}}}$
$\Rightarrow \dfrac{4}{5}={{e}^{\dfrac{-6\pi {{\eta }_{1}}r(40)}{2m}}}$
$\Rightarrow \ln \left( \dfrac{4}{5} \right)=\dfrac{-6\pi {{\eta }_{1}}r(40)}{2m}$ …… (v).
When the pendulum is in carbon dioxide, we can write its amplitude as
$\Rightarrow A={{A}_{0}}{{e}^{\dfrac{-6\pi {{\eta }_{1}}rt}{2m}}}$.
Let the amplitude change from 10cm to 5cm in time t.
$\Rightarrow 5=10{{e}^{\dfrac{-6\pi {{\eta }_{2}}r(t)}{2m}}}$
$\Rightarrow \dfrac{1}{2}={{e}^{\dfrac{-6\pi {{\eta }_{2}}rt}{2m}}}$
$\Rightarrow \ln \left( \dfrac{1}{2} \right)=\dfrac{-6\pi {{\eta }_{2}}rt}{2m}$ …… (vi).
Divide (v) by (vi).
$\Rightarrow \dfrac{\ln \left( \dfrac{4}{5} \right)}{\ln \left( \dfrac{1}{2} \right)}=\dfrac{\dfrac{-6\pi {{\eta }_{1}}r(40)}{2m}}{\dfrac{-6\pi {{\eta }_{2}}rt}{2m}}$
$\Rightarrow \dfrac{2\ln 2-\ln 5}{-\ln 2}=\dfrac{40{{\eta }_{1}}}{t{{\eta }_{2}}}$
It is given that $\dfrac{{{\eta }_{1}}}{{{\eta }_{2}}}=\dfrac{1}{3}$.
$\Rightarrow \dfrac{2\ln 2-\ln 5}{-\ln 2}=\dfrac{40}{3t}$
$\Rightarrow \dfrac{2(0.693)-1.601}{-0.693}=\dfrac{40}{3t}$.
$\Rightarrow 0.31=\dfrac{40}{3t}$
$\Rightarrow t=\dfrac{40}{3(0.31)}=43s$.
This means that the amplitude will reduce from 10cm to 5cm in a time interval of 43s.

Note:
From this question, we get to know that the amplitude of the damped oscillations with respect to time is dependent on the viscosity of the medium in which it is oscillating.
If we place the same bob in different mediums, the rate of decrease in amplitude (also called rate of damping) will be more in that medium which has a higher coefficient of viscosity.