Question

If a simple pendulum has a significant amplitude ( up to a factor of $\dfrac{1}{e}$ of original) only in the period between t = o to t = $\tau$s, then $\tau$ may be called as the average life of the pendulum. When the spherical bob of the simple pendulum suffers a retardation (due to the viscous drag) proportional to its velocity, with b as the constant of proportionality, the average life of the pendulum is (assuming damping is small) in seconds

Answer 1

In the question it is given to us that the pendulum is experiencing a damping force i.e. the viscous drag. Therefore the amplitude of the simple pendulum will gradually decrease with time. First we need to know how the amplitude of the simple pendulum varies with time. Further accordingly we can determine the instant of time i.e. $\tau$ for which the amplitude is equal to $\dfrac{1}{e}$ of original.

Formula used:
$\dfrac{{{d}^{2}}x}{d{{t}^{2}}}+\left( \dfrac{b}{m} \right)\dfrac{dx}{dt}+\dfrac{k}{m}x=0$
$x=A{{e}^{\dfrac{-bt}{2}}}\cos (\omega t+\phi )$

Complete step-by-step answer:
Let us say we have a simple pendulum initially placed in vacuum. Therefore the equation of motion for the pendulum is given by,
$m\dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-kx$
Where ‘m’ is the mass of the bob suspended, $\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$ is the acceleration of the pendulum and ‘k’ is the constant of stiffness. Now if the same pendulum is placed in air such that the bob experiences the viscous force proportional to the velocity i.e. $\dfrac{dx}{dt}$ of the bob and ‘b’ being the constant of proportionality, then the equation of motion is given by
$\begin{aligned} & m\dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-kx-b\dfrac{dx}{dt} \\\ & \Rightarrow m\dfrac{{{d}^{2}}x}{d{{t}^{2}}}+b\dfrac{dx}{dt}+kx=0 \\\ & \therefore \dfrac{{{d}^{2}}x}{d{{t}^{2}}}+\left( \dfrac{b}{m} \right)\dfrac{dx}{dt}+\dfrac{kx}{m}=0 \\\ \end{aligned}$
On solving the above differential equation using the initial conditions we get the equation of motion of the pendulum as,
$x=A{{e}^{\dfrac{-bt}{2}}}\cos (\omega t+\phi )$
Where $\omega =\sqrt{\dfrac{k}{m}-\dfrac{{{b}^{2}}}{4}}$
‘A’ is the maximum amplitude of vibration and $\phi$ is the phase constant of oscillation. If we observe the above equation, the amplitude is a function of time more precisely gradually decreases with time in an exponential manner. The amplitude ‘B’ at any instant of time is given by,
$B=A{{e}^{\dfrac{-bt}{2}}}$
In the question it is given that at time t = $\tau$ the amplitude i.e. $B=\dfrac{1}{e}A$ . hence from the expression for amplitude we get $\tau$ equal to,
$\begin{aligned} & \dfrac{A}{e}=A{{e}^{\dfrac{-b\tau }{2}}} \\\ & \Rightarrow \dfrac{1}{e}={{e}^{\dfrac{-b\tau }{2}}} \\\ & \Rightarrow {{e}^{\dfrac{b\tau }{2}}}=e \\\ & \Rightarrow \dfrac{b\tau }{2}=\log e,\text{ }\because \text{log}e=1 \\\ & \therefore \tau =\dfrac{2}{b} \\\ \end{aligned}$
Therefore the time taken for the pendulum to reach $\dfrac{1}{e}$ of the maximum amplitude is $\dfrac{2}{b}$ .

Note: The drag force in the equation of motion is taken to be negative as it acts opposite to the direction of restoring force. The differential equation is solved by using the initial conditions i.e. at t = 0, x=0. It is also to be noted that the solution of the differential equation obtained can be reduced to many forms depending on the need of answer.