Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If

Q: Why is the decay constant in the formula divided by $2m$?

In the differential equation for damped SHM ($m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$), dividing by $m$ gives the damping term as $\frac{b}{m} \frac{dx}{dt}$. The solution to this differential equation naturally introduces the factor of $2$ in the denominator, resulting in the decay envelope $e^{-bt/2m}$.

Q: Do we need to convert the mass from grams to kilograms?

You can, but it's not strictly necessary here as long as you are consistent. Since the mass is in grams and the damping constant $b$ is in grams per second, the unit 'grams' will cancel out when you calculate the ratio $m/b$, leaving the time purely in seconds.

Q: What is the difference between the decay of amplitude and the decay of energy?

Amplitude decays according to $e^{-bt/2m}$, while mechanical energy, which is proportional to the square of the amplitude ($E \propto A^2$), decays according to $(e^{-bt/2m})^2 = e^{-bt/m}$. This means energy decays twice as fast as the amplitude!

Q: Why did we invert the equation to get $2 = e^{bt/2m}$?

It's a mathematical convenience. We had $1/2 = e^{-bt/2m}$. Inverting both sides (taking the reciprocal) removes the negative sign from the exponent, making the subsequent logarithm step cleaner and less prone to silly sign errors.

Q: What does the term 'decay constant' mean physically in this context?

In this specific context, it refers to the damping constant $b$, which represents the strength of the resistive force (like air drag or fluid viscosity) opposing the motion. A higher $b$ means the oscillations will die out much faster.

Question

Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500 g, decay constant = 20 g/s, then how much time is required for the amplitude of the system to drop to half of its initial value ? (ln 2 = 0.693)

Options:

(A) 34.65 s

(B) 17.32 s

(C) 0.034 s

(D) 15.01 s

Sigma Percentile · Accepted Answer

### The Reality of Friction

In the idealized world of physics, a simple harmonic oscillator swings back and forth forever. But in reality, friction and air resistance always play a role. This is called **Damped Simple Harmonic Motion**.

Imagine a mass-spring system oscillating. Because of the resistive forces, its amplitude doesn't stay constant; it gradually dies out. If you were to plot this motion, you would see a sine wave squeezed inside an exponentially decaying envelope. The rate at which this amplitude shrinks is governed by the damping constant, denoted by $b$.

### The Mathematics of Decay

For a damped oscillator, the amplitude at any given time $t$ is mathematically described by an exponential decay formula:

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

Here, $A_0$ is the initial amplitude, $m$ is the mass of the oscillating body, and $b$ is the damping constant (often referred to as the decay constant in some contexts). The negative sign in the exponent ensures that as time $t$ increases, the amplitude $A(t)$ decreases.

### Finding the Half-Life of Amplitude

The question asks for the exact time when the amplitude drops to exactly half of its initial value. To find this, we set our current amplitude $A(t)$ to $\frac{A_0}{2}$ and substitute it into our master equation:

$$\frac{A_0}{2} = A_0 e^{-\frac{bt}{2m}}$$

We can cleanly cancel $A_0$ from both sides. This leaves us with:

$$\frac{1}{2} = e^{-\frac{bt}{2m}}$$

To get rid of that pesky negative exponent and make our algebra cleaner, we can simply invert both sides (take the reciprocal):

$$2 = e^{\frac{bt}{2m}}$$

Now, to bring the time variable down from the exponent, we take the natural logarithm ($\ln$) on both sides. Since $\ln(e^x) = x$, we get:

$$\ln 2 = \frac{bt}{2m}$$

### The Final Calculation

Let's rearrange this equation to solve for our target variable, time $t$:

$$t = \frac{2m \ln 2}{b}$$

Now, we just plug in the given values. The mass $m$ is $500 	ext{ g}$, the damping constant $b$ is $20 	ext{ g/s}$, and $\ln 2$ is given as $0.693$. Notice that we don't even need to convert grams to kilograms because the unit 'grams' will perfectly cancel out in the ratio $\frac{m}{b}$!

$$t = \frac{2 	imes 500 	imes 0.693}{20}$$

$$t = \frac{1000 	imes 0.693}{20}$$

$$t = 50 	imes 0.693 = 34.65 	ext{ s}$$

And there we have it! It takes exactly $34.65$ seconds for the amplitude to halve.

### A Pro-Tip on Energy Decay

While this question asked about amplitude, JEE often tests your understanding of **energy decay** in damped systems. Remember that the total mechanical energy of an oscillator is proportional to the square of its amplitude ($E \propto A^2$).

If we square the amplitude equation, the factor of $2$ in the denominator of the exponent disappears:

$$E(t) = E_0 e^{-\frac{bt}{m}}$$

This means that the energy decays **twice as fast** as the amplitude! Always read the question carefully to see whether it's asking for the half-life of the amplitude or the half-life of the energy.

Select Answer:

Visualized Solution

Visualizing Damped SHM

The Amplitude Equation

Setting the Condition

Simplifying the Equation

Applying Logarithm

Rearranging and Substituting

Final Calculation

The Way Forward

The Sigma Insight: Free, Forced, and Damped Oscillations

The Reality of Friction

The Mathematics of Decay

Finding the Half-Life of Amplitude

The Final Calculation

A Pro-Tip on Energy Decay

Similar Questions

A particle of mass $m$ is attached to a spring (of spring constant $k$ ) and has a natural angular frequency $ω_{0}$ . An external force $F (t)$ proportional to $cos ω t (ω \neq = ω_{0})$ is applied to the oscillator. The time displacement of the oscillator will be proportional to

An $L$ - $C$ - $R$ circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant $b$ , the correct equivalence would be