If a simple pendulum has significant amplitude (upto a factor of 1/e of original) only in the period

Q: Why is the damping force written as $-mbv$ instead of just $-bv$?

The problem states that the *retardation* (which is deceleration) is proportional to velocity, so $a_{damping} = -bv$. Since Force = mass $\times$ acceleration, the damping force is $m \times (-bv) = -mbv$. It's a subtle but crucial detail!

Q: Where does the formula $A(t) = A_0 e^{-bt/2}$ come from?

It comes from solving the second-order linear differential equation $m\frac{d^2x}{dt^2} + mb\frac{dx}{dt} + kx = 0$. The roots of the characteristic equation have a real part equal to $-\frac{mb}{2m} = -\frac{b}{2}$, which becomes the coefficient in the exponential decay term.

Q: What is the physical significance of dropping to $1/e$?

In any exponential decay process (like $e^{-kt}$), the time $t = 1/k$ is called the time constant or average life. At this time, the value drops to $e^{-1} \approx 0.368$ or 36.8% of its initial value. It's a standard benchmark used across physics to characterize how fast a system loses energy.

Q: Does the initial amplitude $A_0$ affect the average life $\tau$?

Not at all! As you can see in the calculation, $A_0$ cancels out from both sides of the equation. Whether you pull the pendulum back 5 degrees or 10 degrees, the time it takes to drop to $36.8\%$ of that initial angle will be exactly the same.

Q: Why did we ignore the sine/cosine part of the oscillation equation?

The sine/cosine part ($\sin(\omega t + \phi)$) describes the back-and-forth swinging motion. However, the question specifically asks about the *amplitude* (the envelope of the swings), which is governed entirely by the exponential term $A_0 e^{-bt/2}$.

Question

If a simple pendulum has significant amplitude (upto a factor of $1/e$ of original) only in the period between $t = 0$ s to $t = 	au$ s, then $	au$ 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 of proportionality, the average life time of the pendulum (assuming damping is small) in seconds is

Options:

(A) $\frac{0.693}{b}$

(B) $b$

(C) $\frac{1}{b}$

(D) $\frac{2}{b}$

Sigma Percentile · Accepted Answer

### The Reality of the Physical World

Imagine you are standing in a quiet room, watching a simple pendulum swing back and forth. In the idealized world of textbook physics, this pendulum would swing forever. It would never lose energy, and its amplitude would remain perfectly constant.

But we live in the real world.

In reality, the pendulum must push through the air. It experiences air resistance, or what physicists call a **viscous drag**. This drag force constantly saps energy from the system, causing the pendulum's swings to become smaller and smaller until it eventually comes to a complete stop. This beautiful, gradual fading of motion is what we call **Damped Harmonic Motion**.

### Setting Up the Physics

Let's break down the forces acting on the bob of our pendulum.

First, we have the familiar **restoring force**. This is the force that always tries to pull the pendulum back to its lowest point (the equilibrium position). For small angles, this force is proportional to the displacement $x$, and we write it as:

$F_{restoring} = -kx$

Now, let's introduce the star of this problem: the **damping force**. The problem states that the pendulum suffers a retardation (which is a negative acceleration) proportional to its velocity. The constant of proportionality is given as $b$.

So, the retardation is $a_{damping} = -bv$.

Since force is mass times acceleration, the damping force acting on the bob is:

$F_{damping} = m(-bv) = -mbv$

Notice the negative sign! It is crucial. It tells us that the drag force always acts in the *opposite* direction to the pendulum's velocity. If the pendulum is moving right, the drag pushes left.

### The Master Equation

Now, we bring in Sir Isaac Newton. According to Newton's Second Law, the net force on an object equals its mass times its acceleration.

$F_{net} = ma$

Substituting our two forces into this equation, we get:

$ma = -kx - mbv$

Let's bring all the terms to one side to reveal the classic structure of a differential equation:

$ma + mbv + kx = 0$

I know this differential equation might look a bit intimidating, but let's take a breath. This equation is simply a mathematical sentence that says: "The pendulum's inertia, plus the drag of the air, plus the pull of gravity, all balance out perfectly at every instant in time."

### The Exponential Decay

When mathematicians solve this specific differential equation, they find something incredibly elegant. They discover that the pendulum still oscillates, but its amplitude (the maximum distance it swings) shrinks over time.

Specifically, the amplitude follows an **exponential decay** curve. The formula for the amplitude $A(t)$ at any given time $t$ is:

$A(t) = A_0 e^{-\frac{bt}{2}}$

Here, $A_0$ is the initial amplitude (how far you pulled the pendulum back before letting go), and $e$ is Euler's number, the famous mathematical constant approximately equal to 2.718.

The term in the exponent, $-\frac{bt}{2}$, dictates exactly how fast the swings die down. A larger $b$ means more drag, which means the amplitude drops faster.

### The Concept of Average Life

The problem introduces a fascinating concept: the **average life** of the pendulum, denoted by the Greek letter $	au$ (tau).

It defines this average life as the time it takes for the amplitude to drop to a very specific fraction of its original value: exactly $\frac{1}{e}$.

Why $\frac{1}{e}$? In physics and engineering, exponential processes are everywhere—from radioactive decay to discharging capacitors. The time it takes for a quantity to drop to $\frac{1}{e}$ (which is about 37%) of its initial value is a standard, universal way to measure how "fast" the decay is happening. It is a natural benchmark.

So, the problem is giving us a mathematical condition. It says that at time $t = 	au$, the amplitude $A(	au)$ is exactly $\frac{A_0}{e}$.

Let's write that down:

$A(	au) = \frac{A_0}{e}$

### The Final Calculation

We are now in the final stretch. We have our general formula for the amplitude, and we have our specific condition for the average life. Let's merge them!

Substitute $t = 	au$ into our amplitude formula and set it equal to $\frac{A_0}{e}$:

$\frac{A_0}{e} = A_0 e^{-\frac{b	au}{2}}$

This is where the magic happens. Notice how $A_0$ appears on both sides? We can divide both sides by $A_0$, completely eliminating it from the equation. This tells us something profound: the average life of the pendulum does *not* depend on how far you initially pulled it!

We are left with:

$\frac{1}{e} = e^{-\frac{b	au}{2}}$

We can rewrite $\frac{1}{e}$ as $e^{-1}$:

$e^{-1} = e^{-\frac{b	au}{2}}$

Since the bases are the same (both are $e$), their exponents must be perfectly equal. Let's equate them:

$-1 = -\frac{b	au}{2}$

The negative signs cancel out beautifully:

$1 = \frac{b	au}{2}$

Now, we just need to isolate $	au$. Multiply both sides by 2, and divide by $b$:

$	au = \frac{2}{b}$

And there we have it! The average life of the pendulum is exactly $\frac{2}{b}$.

Looking at our options, this matches option (d) perfectly.

This problem is a wonderful journey. It takes us from the physical reality of air resistance, through the rigorous logic of Newton's laws and differential equations, all the way to a clean, elegant algebraic conclusion. Did you get the feel of it?

Select Answer:

Visualized Solution

$D am p e d H a r m o ni c M o t i o n$

$E q u a t i o n o f M o t i o n$

$A m pl i t u d e D ec a y$

$A v er a g e L i f e (τ)$

$S o l v in g f or τ$

$F ina l A n s w er$

The Sigma Insight: Free, Forced, and Damped Oscillations

The Reality of the Physical World

Setting Up the Physics

The Master Equation

The Exponential Decay

The Concept of Average Life

The Final Calculation

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