Given below are some functions of $x$ and $t$ to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all: (a) $y = 2 \cos(3x) \sin(10t)$ (b) $y = 2\sqrt{x - vt}$ (c) $y = 3 \sin(5x - 0.5t) + 4 \cos(5x - 0.5t)$ (d) $y = \cos x \sin t + \cos 2x \sin 2t$

> [!TLDR] > A function represents a **travelling wave** only if it depends on $x,t$ through the single combination $(x\pm vt)$ *and* stays finite everywhere at all times; it represents a **stationary wave** if its spatial and temporal parts separate, $y=f(x)g(t)$. Applying this: (a) is a **stationary wave**; (b) **fails the finite-everywhere test** (it blows up as $x-vt\to\infty$ and is undefined for $x<vt$), so it represents **neither** a travelling nor a stationary wave; (c) is a **travelling wave**; and (d) is a **stationary wave** (a superposition of two stationary waves). ## The two diagnostic forms -- and the hidden third requirement - **Travelling wave:** $y(x,t)=f(x\pm vt)$ -- the shape moves rigidly through space at speed $v$. But this alone isn't sufficient: a *physically acceptable* wave function must also remain **finite for every value of $x$ and $t$** -- an unbounded or undefined function cannot represent a real physical displacement. - **Stationary wave:** $y(x,t)=f(x)\,g(t)$ -- a fixed spatial pattern whose amplitude oscillates in time. ## (a) $y=2\cos(3x)\sin(10t)$ This is a pure product of a function of $x$ alone and a function of $t$ alone -- exactly the $f(x)g(t)$ form. **This is a stationary wave**, with fixed nodes where $\cos(3x)=0$ and antinodes where $|\cos(3x)|=1$. ## (b) $y=2\sqrt{x-vt}$ At first glance, $x$ and $t$ appear only in the single combination $(x-vt)$, which looks like the travelling-wave signature. But check whether it's a *physically valid* wave function: - For $x<vt$, the quantity under the square root is negative, so $y$ is **not even real-valued** there. - For $x-vt\to\infty$, $y=2\sqrt{x-vt}\to\infty$ -- the displacement **grows without bound**, which no real physical medium can do. A genuine wave disturbance must be a finite, single-valued function of $(x\pm vt)$ for all $x$ and $t$ -- exactly the requirement this function violates. It is also not periodic, so it cannot be written in the separable $f(x)g(t)$ stati

Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an ela

Exercises · 14.13

x

(i) a travelling wave,

(ii) a stationary wave or

(iii) none at all:

(a)

y = 2 \cos(3x) \sin(10t)

(b)

y = 2\sqrt{x - vt}

(c)

y = 3 \sin(5x - 0.5t) + 4 \cos(5x - 0.5t)

(d)

y = \cos x \sin t + \cos 2x \sin 2t

Telangana TsbieTextbookSubjective· 3mImportance★★★★★est

33% · 19/58 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 — Wave Type Classification

Wave Type Classification: From Intuition to Precision

Imagine you're standing at the edge of a still pond. You drop a pebble. Ripples spread outward in circles. Now imagine you're holding one end of a long rope tied to a wall. You flick your wrist once — a single hump travels down the rope, hits the wall, and comes back. These are both waves, but they behave differently. Why?

The key difference lies in what is moving versus what is waving.

The Core Intuition

A wave is a disturbance that carries energy from one place to another without permanently moving the medium itself. But the direction of that disturbance relative to the wave's travel direction gives us our first major classification.

Think of a crowd at a stadium doing "the wave." People stand up and sit down (the disturbance moves up-down), but the wave itself travels around the stadium (left-right). The motion of each person is perpendicular to the wave's travel. That's one type.

Now think of a slinky stretched on a table. If you push one end toward the other, a compression travels along the slinky. Each coil moves along the same line as the wave — forward and back. That's the other type.

The Precise Classification

Waves are classified into two fundamental types based on the relationship between the direction of particle displacement (how the medium moves) and the direction of wave propagation (which way the energy travels).

\perp

Transverse Waves

In a transverse wave, the particles of the medium oscillate perpendicular to the direction the wave travels.

Example: Light waves (electromagnetic waves), waves on a string, water waves (surface component), seismic S-waves
Key feature: The wave has crests (high points) and troughs (low points)
Visual: Think of a rope shaken up and down — the rope moves vertically, the wave moves horizontally

Longitudinal Waves

In a longitudinal wave, the particles of the medium oscillate parallel to the direction the wave travels.

Example: Sound waves in air, seismic P-waves, slinky compressions
Key feature: The wave has compressions (regions of high density) and rarefactions (regions of low density)
Visual: Think of a slinky pushed and pulled — coils bunch up and spread out along the same line the wave moves

Note

Some waves are neither purely transverse nor purely longitudinal. Water waves, for instance, have particles moving in circular paths — a combination of both. These are called surface waves or Rayleigh waves in seismology.

Why This Matters

This classification isn't just academic. It determines:

What materials a wave can travel through: Transverse waves (like light) can travel through vacuum. Longitudinal waves (like sound) need a medium. But mechanical transverse waves (like rope waves) also need a medium — the distinction is about how the medium moves, not whether a medium exists. …

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

A function represents a travelling wave only if it depends on $x,t$ through the single combination $(x\pm vt)$ and stays finite everywhere at all times; it represents a stationary wave if its spatial and temporal parts separate, $y=f(x)g(t)$ . Applying this: (a) is a stationary wave; (b) fails the finite-everywhere test (it blows up as $x-vt\to\infty$ and is undefined for $x<vt$ ), so it represents neither a travelling nor a stationary wave; (c) is a travelling wave; and (d) is a stationary wave (a superposition of two stationary waves).

The two diagnostic forms -- and the hidden third requirement

Travelling wave: $y(x,t)=f(x\pm vt)$ -- the shape moves rigidly through space at speed $v$ . But this alone isn't sufficient: a physically acceptable wave function must also remain finite for every value of $x$ and $t$ -- an unbounded or undefined function cannot represent a real physical displacement.
Stationary wave: $y(x,t)=f(x)\,g(t)$ -- a fixed spatial pattern whose amplitude oscillates in time.

(a) $y=2\cos(3x)\sin(10t)$

This is a pure product of a function of $x$ alone and a function of $t$ alone -- exactly the $f(x)g(t)$ form. This is a stationary wave, with fixed nodes where $\cos(3x)=0$ and antinodes where $|\cos(3x)|=1$ .

(b) $y=2\sqrt{x-vt}$

At first glance, $x$ and $t$ appear only in the single combination $(x-vt)$ , which looks like the travelling-wave signature. But check whether it's a physically valid wave function:

For $x<vt$ , the quantity under the square root is negative, so $y$ is not even real-valued there.
For $x-vt\to\infty$ , $y=2\sqrt{x-vt}\to\infty$ -- the displacement grows without bound, which no real physical medium can do.

A genuine wave disturbance must be a finite, single-valued function of $(x\pm vt)$ for all $x$ and $t$ -- exactly the requirement this function violates. It is also not periodic, so it cannot be written in the separable $f(x)g(t)$ stationary-wave form either. This function therefore represents neither a travelling wave nor a stationary wave.

Watch out

Don't stop at "it's a function of $(x-vt)$ , so it must be a travelling wave." That test is necessary but not sufficient -- the function also has to stay finite/bounded and single-valued everywhere. $2\sqrt{x-vt}$ fails that second, equally important condition.

(c) $y=3\sin(5x-0.5t)+4\cos(5x-0.5t)$

Both terms share the identical argument $\theta=5x-0.5t$ . Using $A\sin\theta+B\cos\theta=R\sin(\theta+\alpha)$ with $R=\sqrt{A^2+B^2}$ :

$R = \sqrt{3^2+4^2} = 5, \qquad \tan\alpha = \frac43$ …

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.

TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
$\rightarrow$
›Reveal solutionSolution

Match each radiation to the use that exploits its wavelength: X-rays→crystal diffraction, UV→forensic fingerprints, radio→TV transmission, IR→remote controls. That is A→R, B→Q, C→S, D→P, i.e. option (B).

The concept first

The electromagnetic spectrum is one continuous family — all of it is $\vec E$ and $\vec B$ oscillating and travelling at $c$ . What changes across the spectrum is the wavelength $\lambda$ and hence the photon energy $E=h\nu=hc/\lambda$ . Every practical application is chosen because $\lambda$ (or $E$ ) fits the physical scale of the problem. Learn that logic and you never have to memorise the table.

Step-by-step

1. X-rays $\to$ crystal structure study (R).

X-rays have $\lambda\sim10^{-10}\,\text{m}=1\ \text{Å}$ , which is precisely the spacing between atomic planes in a crystal. A wave diffracts strongly only when the obstacle spacing is comparable to $\lambda$ — Bragg's law,
$2d\sin\theta=n\lambda .$
So X-rays (and only X-rays) give a usable diffraction pattern from a crystal lattice. This is how DNA's structure and countless mineral structures were found.

2. UV rays $\to$ finger prints in forensic labs (Q).

UV photons ( $E\approx3$ – $10$ eV) are energetic enough to excite electrons in the organic residues (oils, amino acids) left behind by a fingertip, and in the fluorescent powders dusted on a surface. These re-emit visible light — the print glows against its background. Visible light cannot do this; X-rays would be far too penetrating.

3. Radio waves $\to$ TV communication system (S). …
Start your 14-day free trial to see the full solution →
TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$Q. Among the following statements, the correct statement for a wave is (A) Transverse waves cannot propagate through all media (B) Longitudinal waves can propagate through solids only (C) Transverse waves can propagate through solids (D) Longitudinal waves can propagate through vacuum$
›Reveal solutionSolution
The key idea is that wave propagation depends on the medium’s ability to support the required deformation: transverse waves need shear rigidity (solids only), while longitudinal waves need bulk elasticity (solids, liquids, gases). The correct statement is (C).

The question tests your understanding of how mechanical waves travel through different media. The crucial distinction is between transverse waves (where particles oscillate perpendicular to the wave’s direction) and longitudinal waves (where particles oscillate parallel to the wave’s direction). Each type requires a specific kind of elastic restoring force from the medium.

Transverse waves rely on shear stress — the medium must resist being twisted or bent sideways.

Solids have a fixed shape and can provide shear restoring forces, so transverse waves (like waves on a string or seismic S-waves) can travel through solids.

Fluids (liquids and gases) cannot sustain shear stress; they flow to eliminate sideways deformation. Therefore, transverse mechanical waves cannot propagate through liquids or gases.

This immediately tells us statement (C) is true: “Transverse waves can propagate through solids.”

It also tells us statement (A) is false: transverse waves cannot propagate through all media (they fail in fluids), but the phrasing “cannot propagate through all media” is ambiguous — it actually means “there exist media through which they cannot propagate,” which is true, but the intended meaning here is likely “they cannot propagate through any medium,” which is false. In standard MCQs, (A) is considered incorrect because transverse waves can propagate through solids.

Longitudinal waves rely on bulk compression — the medium must be able to be squeezed and expanded.

Solids, liquids, and gases all resist compression (they have bulk modulus), so longitudinal waves (sound waves, seismic P-waves) travel through all three. …
Start your 14-day free trial to see the full solution →
TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
$Q. Consider following statements. A) Longitudinal waves need bulk modulus of elasticity and therefore can propagate in all media. B) Progressive wave is a wave that moves from one point of medium to another. C) In a wave, energy and matter are transferred from one point to another. Choose the correct option. (A) A, B, C are correct (B) only B and C are correct (C) only A and B are correct (D) only A and C are correct$
›Reveal solutionSolution

Longitudinal waves require bulk modulus and propagate in all media; progressive waves transfer energy by moving through the medium. Only statements A and B are correct because waves transfer energy, not matter.

The question tests your understanding of wave fundamentals: what physical properties enable wave propagation, what defines a progressive wave, and the crucial distinction between energy and matter transport.

Longitudinal waves compress and expand the medium along the direction of propagation. This compression requires the medium to resist volume changes, which is quantified by the bulk modulus of elasticity $K$ . Solids, liquids, and gases all possess bulk modulus (though gases have the smallest values), so longitudinal waves—sound being the classic example—can indeed propagate through all three states of matter. Statement A captures this correctly.

A progressive wave (or traveling wave) is one that carries a disturbance from one location to another through the medium. The wave profile moves spatially, described by equations like $y = A \sin(kx - \omega t)$ , where the phase $(kx - \omega t)$ advances in space and time. Statement B is a straightforward, correct definition.

The critical conceptual point lies in statement C. Waves are mechanisms of energy transfer, not mass transfer. When a wave passes through a medium, individual particles oscillate about their equilibrium positions—they do not travel with the wave. A water wave moves across a pond, but the water molecules themselves bob up and down; they don't flow horizontally with the wave. Energy propagates, matter stays local.

Let me evaluate each statement: …
Start your 14-day free trial to see the full solution →
TG EAPCET 2021Set ap-2021-08-09-AN1 markMCQ
$[l] = [l_0] = L$
›Reveal solutionSolution
The key idea is that arguments of transcendental functions (sin, cos, log) must be dimensionless. Only option (A) violates this by taking the sine of a dimensioned quantity, making it dimensionally inconsistent. The correct option is (A).

Concept & Intuition

In physics, every equation must be dimensionally consistent: the dimensions on both sides must match. A subtle but crucial rule is that transcendental functions (trigonometric, logarithmic, exponential, etc.) can only act on pure numbers — quantities with no physical dimension. If you see $\sin(m)$ where $m$ has dimensions of mass (M), that’s a red flag. The argument inside a sine, cosine, or log must be dimensionless. We’ll check each option by examining the arguments of these functions.

Option (A): $l = l_0 \dfrac{\sin(m)}{\sin(m_0)}$

Here $l$ and $l_0$ have dimension $L$ , so the right side must also have dimension $L$ . The ratio $\sin(m)/\sin(m_0)$ should be dimensionless.

But $m$ and $m_0$ have dimension $M$ (mass). The sine function requires a dimensionless argument — you cannot take the sine of a mass.

Conclusion: This is dimensionally inconsistent because $\sin(m)$ is meaningless.

Already we suspect (A) is the answer, but let’s verify the others.

Option (B): $l = l_0 \dfrac{\sin(m/m_0)}{\cos(m/m_0)}$

The arguments $m/m_0$ are ratios of masses, so they are dimensionless (M/M = 1).

$\sin$ and $\cos$ of a dimensionless number are fine. The ratio $\sin/\cos$ is also dimensionless.

Multiplying by $l_0$ (dimension L) gives L on the right, matching $l$ on the left.

Conclusion: Dimensionally consistent.

Option (C): $l = l_0 \dfrac{\log(m/m_0)}{\sin(m/m_0)}$

Again $m/m_0$ is dimensionless, so $\log$ and $\sin$ are valid. The ratio is dimensionless.

Right side has dimension L, left side L.

Conclusion: Dimensionally consistent. …
Start your 14-day free trial to see the full solution →
TG EAPCET 2021Set ap-2021-08-09-AN1 markMCQ
$Q. In frequency modulated wave. (A) both frequency and amplitude vary with time (B) both frequency and amplitude are constant (C) frequency varies with time (D) amplitude varies with time$
›Reveal solutionSolution
In frequency modulation (FM), the amplitude of the carrier wave stays constant while its frequency changes in step with the modulating signal. So only frequency varies with time — the correct answer is (C).

The Core Concept

Frequency modulation (FM) is a method of encoding information in a carrier wave by varying its instantaneous frequency according to the amplitude of the input signal. The key insight: the amplitude of the FM wave is deliberately kept fixed — this is what distinguishes FM from AM (amplitude modulation). Think of it like a singer holding a steady loudness (constant amplitude) but changing pitch (frequency) to convey a melody.

Step-by-Step Reasoning

What is modulated?

In FM, the frequency of the carrier wave is varied around a central (carrier) frequency. The amount of variation is proportional to the instantaneous amplitude of the modulating signal (e.g., audio). So the frequency is not constant — it changes continuously with time.

What stays constant?

The amplitude of the FM wave is kept constant. This is a deliberate design choice: by not varying the amplitude, FM is immune to amplitude noise (like static from lightning or electrical interference). The transmitted power is also constant, which simplifies amplifier design.

Mathematical representation

An FM wave can be written as:

$s(t) = A_c \cos\left(2\pi f_c t + 2\pi k_f \int_0^t m(\tau)\,d\tau\right)$
Here, $A_c$ is the constant amplitude, $f_c$ is the carrier frequency, and $m(t)$ is the modulating signal. The instantaneous frequency is $f_c + k_f m(t)$ , which clearly varies with time. The amplitude $A_c$ does not depend on time.

Eliminating the wrong options …
Start your 14-day free trial to see the full solution →
TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
$Q. Which of the following is not a transverse wave? (A) Light waves (B) Sound waves (C) Waves on a violin string (D) Water waves$
›Reveal solutionSolution
The key idea is that transverse waves require a medium that can sustain shear stress, while longitudinal waves involve compression and rarefaction. Sound waves are longitudinal, not transverse, so the answer is (B).

The concept here is the classification of waves based on the direction of particle displacement relative to the wave's propagation. In a transverse wave, particles of the medium oscillate perpendicular to the direction the wave travels. In a longitudinal wave, particles oscillate parallel to the wave's direction. The critical physical distinction is that transverse waves need a medium with shear rigidity — the ability to resist sideways deformation — while longitudinal waves only need bulk elasticity (compressibility). This is why sound, which travels through fluids (gases and liquids) that cannot support shear, is always longitudinal.

Let’s examine each option:

Light waves (A) — These are electromagnetic waves. They consist of oscillating electric and magnetic fields, and the field oscillations are perpendicular to the direction of propagation. Light does not require a medium at all, but by definition it is transverse. So this is a transverse wave.

Sound waves (B) — Sound propagates through air, water, or solids as a series of compressions and rarefactions. The air molecules move back and forth along the direction the sound travels. In fluids, there is no shear strength, so transverse waves cannot exist. Even in solids, sound can have a longitudinal component, but the primary mode in air is longitudinal. Hence, sound is not a transverse wave. …
Start your 14-day free trial to see the full solution →

←Previous question14.12 — (i) For the wave on a string described in Exercise 14.11, do all the p…→Next question14.14 — A wire stretched between two rigid supports vibrates in its fundamenta…

Q.Given below are some functions of xxx and ttt to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent

The two diagnostic forms -- and the hidden third requirement

(a) y=2cos⁡(3x)sin⁡(10t)y=2\cos(3x)\sin(10t)y=2cos(3x)sin(10t)

(b) y=2x−vty=2\sqrt{x-vt}y=2x−vt​

(c) y=3sin⁡(5x−0.5t)+4cos⁡(5x−0.5t)y=3\sin(5x-0.5t)+4\cos(5x-0.5t)y=3sin(5x−0.5t)+4cos(5x−0.5t)

🎓Unlock everything free for 14 days

Q.Given below are some functions of $x$ and $t$ to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent

(a) $y=2\cos(3x)\sin(10t)$

(b) $y=2\sqrt{x-vt}$

(c) $y=3\sin(5x-0.5t)+4\cos(5x-0.5t)$

Unlock everything free for 14 days