A pendulum oscillates in air, so it experiences air resistance (a damping force). Consider how its total mecha

NCERT Exemplar · Q13

E

(a) A curve that oscillates about the time axis with steadily decreasing amplitude (crossing zero repeatedly).

(b) A curve that starts at a high value and falls to zero at a finite time along a concave arc.

(d) A sinusoidal curve of roughly constant amplitude oscillating about zero.

Andhra Pradesh BieapMCQ· 1mImportance★★★★★est

58% · 48/83 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Damped Oscillations

Damped Oscillations: When Motion Fades

Imagine pushing a child on a swing. You give one big push, then step back. The swing goes high, then lower, then lower still, until eventually it stops. That is a damped oscillation in everyday life. The swing wants to keep swinging forever — that would be an ideal, undamped oscillation — but something is stealing its energy. Air resistance, friction at the pivot, even the slight bending of the ropes all act as a brake.

The key intuition: the system still oscillates, but each swing is a little smaller than the last. The amplitude does not drop suddenly; it shrinks in a smooth, predictable way — exponentially.

The Physics: Where Does the Energy Go?

In an ideal oscillator (like a mass on a spring with no friction), the total mechanical energy is constant. Kinetic energy converts to potential energy and back, forever. The equation of motion is:

m\frac{d^2x}{dt^2} + kx = 0

where $m$ is mass, $k$ is spring constant, and $x$ is displacement.

Now add a resistive force. The simplest model is a force proportional to velocity, like air drag at low speeds or the friction in a dashpot (a piston in oil). That force is:

F_{\text{damping}} = -b v = -b \frac{dx}{dt}

where $b$ is the damping coefficient — a positive number that measures how strong the resistive force is. The minus sign means the force always opposes the motion.

Newton's second law then becomes:

m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx = 0

That is the damped harmonic oscillator equation. It is the precise statement.

The Solution: Exponential Decay of Amplitude

The solution to this differential equation depends on how strong the damping is. For the most common case — underdamping — the system still oscillates, and the displacement is:

x(t) = A_0 e^{-\frac{b}{2m}t} \cos(\omega' t + \phi)

Here is what each piece means:

$A_0$ is the initial amplitude.
$e^{-\frac{b}{2m}t}$ is the exponential decay factor. As time $t$ increases, this factor shrinks from 1 toward 0. The quantity $\frac{b}{2m}$ is often written as $\gamma$ (the damping constant) or $\beta$ .
$\cos(\omega' t + \phi)$ is the oscillatory part, with a new angular frequency $\omega'$ that is slightly less than the natural frequency $\omega_0 = \sqrt{k/m}$ :

\omega' = \sqrt{\omega_0^2 - \left(\frac{b}{2m}\right)^2}

Important

The amplitude of a damped oscillation decays as $A(t) = A_0 e^{-\frac{b}{2m}t}$ . The energy, which is proportional to amplitude squared, decays as $E(t) = E_0 e^{-\frac{b}{m}t}$ .

Three Regimes of Damping

Not all damped systems oscillate. The value of $b$ relative to the critical value $b_c = 2\sqrt{km}$ decides the behaviour:

Regime	Condition	Behaviour
Underdamped	$b < 2\sqrt{km}$	Oscillates with decaying amplitude
Critically damped	$b = 2\sqrt{km}$	Returns to equilibrium fastest, no oscillation
Overdamped	$b > 2\sqrt{km}$	Returns slowly, no oscillation

Note

Critical damping is the sweet spot for things like door closers and car shock absorbers — you want the system to settle to zero as quickly as possible without bouncing.

Why Exponential? The Intuition …

Start your 14-day free trial to unlock the full solution →

Air resistance continuously drains energy from the swinging pendulum, so its total mechanical energy only decreases with time — smoothly and gradually, never oscillating and never negative. This is the exponential-type decay of option (C).

Reasoning

The kinetic and potential energies individually oscillate as the bob swings, but their sum (total mechanical energy) is what the question asks about. A resistive (damping) force does negative work every cycle, so:

\frac{dE}{dt} < 0 \quad\text{at all times}

The energy therefore decreases monotonically, tending to zero as the oscillations die out — an exponential-type decay.

Why the others are wrong …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
$Q. The amplitude of a damped harmonic oscilator becomes 50% of its initial value in a time of 12 s. If the amplitude of the oscillator at a time of 36 s is x% of its initial amplitude, then the value of x is (A) 25 (B) 12.5 (C) 37.5 (D) 8$
›Reveal solutionSolution
Because damped-amplitude decay is exponential, halving every 12 s means after three such intervals (36 s) the amplitude is $(1/2)^3=12.5\%$ of the initial value.

Concept and Intuition

A damped oscillator's amplitude decays as $A(t)=A_0 e^{-\gamma t}$ — an exponential decay, exactly like radioactive decay with a 'half-life'. If the amplitude drops to 50% in 12 s, then in every subsequent 12 s interval it again drops to 50% of whatever it was at the start of that interval — the fractions multiply, they don't add.

Step-by-Step Solution

Given: $A(12) = 0.5\,A_0 \Rightarrow e^{-12\gamma}=0.5$ .

$36\,\text{s} = 3\times12\,\text{s}$ , so $A(36) = A_0 e^{-36\gamma} = A_0\left(e^{-12\gamma}\right)^3 = A_0(0.5)^3$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
$x(t) = \exp(-0.2t)\cos(3.2t + \phi)$
›Reveal solutionSolution
The exponential envelope $e^{-0.2t}$ is the amplitude of a damped oscillator; matching it to $e^{-1.2}$ directly gives $t=6$ s.

Concept and Intuition

For a damped oscillator $x(t)=A_0e^{-bt}\cos(\omega' t+\phi)$ , the oscillating part just tells you the phase of oscillation, while the slowly-decaying envelope $A_0e^{-bt}$ is the instantaneous amplitude. Comparing envelopes at two different times only ever requires equating the exponents.

Step-by-Step Solution

Given $x(t)=e^{-0.2t}\cos(3.2t+\phi)$ , so amplitude at time $t$ is $A(t)=e^{-0.2t}$ (with $A(0)=1$ ).

We want $A(t)=\dfrac{1}{e^{1.2}}=e^{-1.2}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
$\dfrac{1}{e}$
›Reveal solutionSolution
Damped-oscillator amplitude decays exponentially; each equal time interval multiplies the amplitude by the same factor. Since the first 2 s gives a factor $1/e$ , the next 2 s gives another factor of $1/e$ , so after 4 s total the amplitude is $A/e^2$ .

Concept and Intuition

Exponential decay has the property that equal time intervals always produce equal ratios of decay, not equal absolute drops. So if the amplitude falls to $1/e$ of its value in the first 2 seconds, it will fall to $1/e$ of whatever it was at the start of the next 2-second interval — compounding to $1/e \times 1/e = 1/e^2$ of the original.

Step-by-Step Solution

Amplitude of a damped oscillator: $A(t) = A\,e^{-\gamma t}$ , where $A$ is the initial amplitude.

Given: $A(2) = A/e \Rightarrow e^{-2\gamma} = e^{-1} \Rightarrow \gamma = \tfrac12\ \text{s}^{-1}$ .

Amplitude at $t=4$ s (i.e. after the next 2 s beyond the first 2 s): $A(4) = A\,e^{-\gamma\times4} = A\,e^{-2}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
$A_1$
›Reveal solutionSolution
Exponential amplitude decay in damped SHM makes $A_1$ the geometric mean of $A_0$ and $A_2$ (times spaced 2s apart, i.e. equal intervals), because the decay is multiplicative, not additive.

Concept and Intuition

A damped harmonic oscillator has amplitude $A(t)=A_0e^{-kt}$ where $k$ is the damping constant. Because the decay is exponential, equal time intervals correspond to equal multiplicative factors — this is the hallmark of geometric (not arithmetic) progression in time. Since $t=2\,\mathrm s$ and $t=4\,\mathrm s$ are symmetric about $t=2\,\mathrm s$ measured from $t=0$ in log-amplitude space is not quite it — rather, $A_1$ at $t=2$ s sits exactly halfway (in exponent) between $A_0$ at $t=0$ and $A_2$ at $t=4$ s, so $A_1$ is the geometric mean of $A_0$ and $A_2$ .

Step-by-Step Solution

Amplitude law: $A(t)=A_0e^{-kt}$ .

At $t=2\,\mathrm s$ : $A_1=A_0e^{-2k}$ .

At $t=4\,\mathrm s$ : $A_2=A_0e^{-4k}$ .

Square $A_1$ : $A_1^2=A_0^2e^{-4k}$ .

Rewrite using $A_2$ : $A_0^2e^{-4k}=A_0\left(A_0e^{-4k}\right)=A_0A_2$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
$\dfrac{1}{5}\log_e 3\text{ s}^{-1}$
›Reveal solutionSolution
This tests the exponential energy-decay law of a damped oscillator,

$E(t)=E_0e^{-bt/m}$ . Answer: $\dfrac{1}{15}\log_e3\ \text{s}^{-1}$ .

Concept and Intuition

In damped SHM, the amplitude decays exponentially as $e^{-bt/2m}$ , where $b$ is the

damping constant. Since energy is proportional to amplitude squared, the energy decays

twice as fast in the exponent: $E(t)=E_0e^{-bt/m}$ . Given the energy at two times, we

can extract $b/m$ directly from the ratio, without needing $b$ and $m$ separately.

Step-by-Step Solution

$E(t)=E_0e^{-bt/m}$ , with $E_0=45$ J at $t=0$ .

Time period $=1$ s, so after 15 oscillations, $t=15\times1=15$ s, and $E=15$ J.

$15=45e^{-15b/m}\Rightarrow e^{-15b/m}=\dfrac{15}{45}=\dfrac13$

Take natural log: $-\dfrac{15b}{m}=\ln\dfrac13=-\ln3 \Rightarrow \dfrac{15b}{m}=\ln3$ …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
$Q. Which of the following statements regarding damping force of a damped oscillator is NOT correct? (A) Damping force depends on the nature of the surrounding medium. (B) Damping force is generally proportional to the velocity of the body making oscillations. (C) Damping force acts in the direction of the velocity of the body. (D) Ratio of the damping force and velocity of the body depends on the size and shape of the body.$
›Reveal solutionSolution
This tests the basic nature of a damping (viscous drag) force: it always opposes

motion. Answer: option (C) is the false statement.

Concept and Intuition

Damping arises from resistive interactions with the surrounding medium (air, liquid,

internal friction), and by its very physical origin it always acts to slow the body

down — i.e. opposite to the direction of motion. A force "in the direction of

velocity" would speed the oscillator up, which is the opposite of damping.

Step-by-Step Solution

Recall the standard damping force model: $F_d = -b v$ , where $b$ is the damping constant and $v$ is the velocity — the negative sign shows it opposes velocity.

Check each statement:

(A) Depends on the medium — true (e.g. air vs. water vs. vacuum).

(B) Proportional to velocity — true, for the standard linear-damping model.

(C) Acts in the direction of velocity — false; it acts opposite to velocity. …
Start your 14-day free trial to see the full solution →

←Previous questionQ12 — The Earth moves once around the Sun in an elliptical orbit. Consider h…→Next questionQ14 — A mass of 5 kg is moving along a circular path of radius 1 m. If the m…

Q.A pendulum oscillates in air, so it experiences air resistance (a damping force). Consider how its total mechanical energy $E$ changes with time. Which of the following $E$ -versus-time curves correctly represents this variation?

Concept understanding — Damped Oscillations

Damped Oscillations: When Motion Fades

The Physics: Where Does the Energy Go?

The Solution: Exponential Decay of Amplitude

Three Regimes of Damping

Why Exponential? The Intuition …

Reasoning

Why the others are wrong …

Concept: Damped Oscillations and Non-Conservative Forces

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.A pendulum oscillates in air, so it experiences air resistance (a damping force). Consider how its total mechanical energy EEE changes with time. Which of the following EEE-versus-time curves correctly represents this variation?

Reasoning

Why the others are wrong …

🎓Unlock everything free for 14 days

Q.A pendulum oscillates in air, so it experiences air resistance (a damping force). Consider how its total mechanical energy $E$ changes with time. Which of the following $E$ -versus-time curves correctly represents this variation?

Unlock everything free for 14 days