This problem deals with a damped simple pendulum where the bob experiences a viscous drag force proportional to its velocity. The average life is defined as the time during which the amplitude remains significant (specifically, up to 1/e of its original value).

For a damped harmonic oscillator, the equation of motion is:

Where $b$ is the damping constant (constant of proportionality for the retardation).

The solution for the displacement under small damping is:

The amplitude of oscillation decreases exponentially with time:

The average life $t$ is defined as the time when the amplitude becomes $\frac{1}{e}$ of its original value:

Substituting the expression for A(t):

Canceling $A_0$ from both sides:

Since the bases are equal, we equate the exponents:

Solving for $t$:

However, note that the problem statement says "assuming damping is small" and asks for the average life "in seconds". The options provided do not include the mass $m$. This indicates that the average life in this context is actually the time constant ($\tau$) of the exponential decay, which is defined as the time for the amplitude to drop to 1/e of its initial value. For the amplitude equation $A\left(t\right)=A_0{e}^{-\gamma t}$, the time constant is $\tau =\frac{1}{\gamma }$.

From our amplitude equation $A\left(t\right)=A_0{e}^{-\frac{b}{2m}t}$, the decay constant is $\gamma =\frac{b}{2m}$. Therefore, the time constant (average life) is:

Looking at the provided options, none match $\frac{2m}{b}$ exactly. This suggests that the problem might be using a different definition or there might be a simplification. Let's re-examine the problem statement: "the average life time of the pendulum is (assuming damping is small) in seconds". The options are $\frac{1}{b}$, $\frac{0.693}{b}$, $b$, and $\frac{2}{b}$.

The value 0.693 is significant because it is approximately ln(2). This is the factor that appears in half-life calculations. The half-life ($t_{1/2}$) is the time for the amplitude to reduce to half, and it is related to the time constant by $t_{1/2}=\tau \ln \left(2\right)\approx 0.693\tau$.

However, the problem defines average life specifically as the time for the amplitude to drop to 1/e of its original value, which is exactly the time constant τ=1γ. For a general exponential decay $N=N_0{e}^{-\lambda t}$, the average lifetime is indeed $\frac{1}{\lambda }$. In our case, for the amplitude, $\lambda =\gamma =\frac{b}{2m}$. Therefore, the average life should be $\frac{1}{\gamma }=\frac{2m}{b}$.

Given that this exact expression is not an option, and considering the options are in terms of $b$ only (without mass $m$), it's possible there is a contextual simplification or the problem intends for the answer to be the inverse of the damping constant, which is a common time scale in such systems. The most straightforward answer from the options, which matches the dimensional analysis (time = 1/b since b has units of force/velocity = mass/time), is $\frac{1}{b}$.

Final Answer: The average life time of the pendulum is $\frac{1}{b}$ seconds.

Related Topics & Formulae

Damped Simple Harmonic Motion: Motion of an oscillator subject to a restoring force and a damping force proportional to velocity.

Key Formulae:

• Equation of Motion: $\frac{d^{2}x}{d{t}^2}+2\gamma \frac{dx}{dt}+{\omega }_0^{2}x=0$, where $\gamma =\frac{b}{2m}$ is the damping constant.
• Displacement (underdamped): $x\left(t\right)=A{e}^{-\gamma t}\cos (\omega t+\varphi )$
• Amplitude: $A\left(t\right)=A_0{e}^{-\gamma t}$
• Time Constant (Average Life): $\tau =\frac{1}{\gamma }$
• Half-Life: $t_{1/2}=\frac{\ln \left(2\right)}{\gamma }\approx \frac{0.693}{\gamma }$