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

Options

A. `(0.693)/(b)`
B. b
C. `(1)/(b)`
D. `(2)/(b)`

Answer

To solve the problem, we need to analyze the motion of a damped simple pendulum and determine the average life time, denoted as \( \tau \), when the amplitude of the pendulum decreases to \( \frac{1}{e} \) of its original amplitude.

### Step-by-Step Solution:

1. **Understanding Damped Motion**:
- A simple pendulum experiences a restoring force due to gravity, which is proportional to the displacement from the mean position. When damping is present (due to viscous drag), the motion becomes a damped harmonic motion.
- The equation of motion for a damped harmonic oscillator can be written as:
\[ m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0 \]
- Here, \( m \) is the mass of the bob, \( b \) is the damping constant, \( k \) is the spring constant (related to the pendulum), and \( x \) is the displacement.

2. **Formulating the Damped Motion Equation**:
- Rearranging the equation gives:
\[ \frac{d^2x}{dt^2} + \frac{b}{m} \frac{dx}{dt} + \frac{k}{m} x = 0 \]
- This is a second-order linear differential equation.

3. **Finding the Solution**:
- The general solution for the displacement \( x(t) \) in damped harmonic motion is given by:
\[ x(t) = A_0 e^{-\frac{b}{2m} t} \cos(\omega_d t + \phi) \]
- Here, \( A_0 \) is the initial amplitude, \( \omega_d = \sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^2} \) is the damped angular frequency, and \( \phi \) is the phase constant.

4. **Amplitude Decay**:
- The amplitude \( A(t) \) at time \( t \) can be expressed as:
\[ A(t) = A_0 e^{-\frac{b}{2m} t} \]
- We need to find the time \( \tau \) when the amplitude decreases to \( \frac{A_0}{e} \):
\[ A(\tau) = \frac{A_0}{e} \]

5. **Setting Up the Equation**:
- Setting the amplitude equation equal to \( \frac{A_0}{e} \):
\[ A_0 e^{-\frac{b}{2m} \tau} = \frac{A_0}{e} \]
- Dividing both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ e^{-\frac{b}{2m} \tau} = \frac{1}{e} \]

6. **Solving for \( \tau \)**:
- Taking the natural logarithm of both sides:
\[ -\frac{b}{2m} \tau = -1 \]
- Rearranging gives:
\[ \tau = \frac{2m}{b} \]

7. **Final Expression**:
- Since we are asked for the average life time of the pendulum in seconds, we can express it as:
\[ \tau = \frac{2}{b} \]
- This is the average life time of the pendulum under the given conditions.

### Conclusion:
The average life time \( \tau \) of the pendulum is given by:
\[ \tau = \frac{2}{b} \text{ seconds} \]