If a simple pendulum has significant amplitude (up to a factor of 1/e of original) only in the period between t=0 to t=τs, then τ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with b' as the constant proportionality, the average life time of the pendulum in seconds (assuming damping is small) in second is?

Consider the free body diagram shown in the figure. For the equilibrium in the normal direction, we have T=mgcosθ In the tangential direction, mLθ'' = -mgsinθ -mbv For small angle θ, we have sinθ ≈ θ and v=Lθ̇ Thus, mLθ'' + mbLθ̇ + mgθ = 0 Solution to the above differential equation is θ(t) = θ0e^(-bt/2)sin(ωdt+φ) Let average lifetime be τ. Amplitude at t=0 is θ0 and amplitude at t=τ is θ0e^(-bτ/2) But the ratio is equal to 1/e Thus, e^(-bτ/2) = e⁻¹ ⇒ τ = 2/b