Exercises — Waves

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14.12 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.1
A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If a transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Hint. Find the linear mass density mu first, then the wave speed v = sqrt(T/mu), and divide the string's length by that speed.
Step 1 — Linear mass density. mu = mass/length = 2.50/20.0 = 0.125 kg/m.

Step 2 — Wave speed. v = sqrt(T/mu) = sqrt(200/0.125) = sqrt(1600) = 40 m/s.

Step 3 — Time to travel the string's length. time = L/v = 20.0/40 = 0.5 s.

✦ The disturbance reaches the other end in 0.5 s, since the jerk travels as a transverse wave whose speed depends only on the string's tension and mass per unit length, not on how sharply it was struck.
14.23 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.2
A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top, given the speed of sound in air is 340 m/s? (g = 9.8 m/s^2.)
Hint. Two separate time intervals add up here: the time for the stone to fall (using kinematics), plus the time for the sound of the splash to travel back up (distance over speed of sound).
Step 1 — Time for the stone to fall. h = (1/2)g t1^2, so t1 = sqrt(2h/g) = sqrt(2 x 300/9.8) = sqrt(61.22) = 7.82 s.

Step 2 — Time for sound to travel back up. t2 = h/v_sound = 300/340 = 0.88 s.

Step 3 — Total time. t_total = t1 + t2 = 7.82 + 0.88 = 8.71 s.

✦ The splash is heard about 8.71 s after the stone is dropped, since the fall and the sound's return trip are two separate, sequential time intervals that must be added, not averaged or compared.
14.33 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.3
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20 degrees C, 343 m/s?
Hint. Find the linear mass density first, then rearrange v = sqrt(T/mu) to solve for T directly.
Step 1 — Find the linear mass density from the given mass and length. mu = 2.10/12.0 = 0.175 kg/m.

Step 2 — Solve for tension. v = sqrt(T/mu), so T = v^2 x mu = 343^2 x 0.175 = 117649 x 0.175 = 20589 N.

✦ The wire must be under a tension of about 2.06 x 10^4 N, since matching a wave speed as high as the speed of sound requires a correspondingly large tension for this particular mass density.
14.43 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.4
Use the formula v = sqrt(gamma P/rho) to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.
Hint. Rewrite the formula using the ideal gas law so that P and rho are replaced by quantities that reveal what the speed actually depends on.
(a) Independent of pressure: from the ideal gas law, rho = PM/(RT) at a given temperature, so P/rho = RT/M — a ratio that does NOT depend on P itself, since increasing P also increases rho proportionally at fixed T. So v = sqrt(gamma RT/M) has no leftover dependence on pressure.

(b) Increases with temperature: from the same substitution, v = sqrt(gamma RT/M) shows v is directly proportional to the square root of absolute temperature T — as T rises, v rises with it.

(c) Increases with humidity: moist air has a lower average molar mass M than dry air, since water vapour molecules (M = 18) are lighter than the average of the N2 and O2 molecules they partially displace (M around 29). Since v is inversely proportional to sqrt(M), a lower effective M means a higher speed of sound.

✦ All three results trace back to rewriting v = sqrt(gamma P/rho) as v = sqrt(gamma RT/M) using the ideal gas law, which reveals that the speed genuinely depends only on temperature and molar mass, not on pressure or density individually.
14.54 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.5
A travelling wave in one dimension is represented by a function y = f(x,t) where x and t must appear in the combination x - vt or x + vt, i.e. y = f(x plus-or-minus vt). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave: (a) (x - vt)^2, (b) log[(x + vt)/x0], (c) 1/(x + vt).
Hint. A genuine physical wave must stay finite (bounded) everywhere, for all x and t — check whether each function blows up anywhere, either at a specific point or as x, t grow large.
The converse is NOT true. Depending on x and t only through the combination x plus-or-minus vt is necessary for a travelling wave, but not sufficient — a physically sensible wave disturbance must also remain FINITE everywhere.

(a) (x - vt)^2 does not blow up at any finite x, t, but it grows without bound as x or t becomes large, and it never oscillates or localises — it does NOT represent a physical travelling wave.

(b) log[(x + vt)/x0] diverges to negative infinity at x + vt = 0, and to positive infinity as x + vt grows large — it does NOT represent a physical travelling wave.

(c) 1/(x + vt) diverges to infinity at x + vt = 0 — it does NOT represent a physical travelling wave.

✦ None of the three qualifies, since each becomes infinite somewhere (either at a specific point or as the argument grows large), and a genuine wave disturbance must stay finite at every position and every instant of time — the combination-of-variables test alone is not enough to guarantee that.
14.63 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.6
A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m/s and in water 1486 m/s.
Hint. Frequency is set by the source and stays the same on both reflection and transmission — only wavelength changes, since it depends on the speed in whichever medium the wave is currently in.
Frequency stays fixed at f = 1000 kHz = 1.0 x 10^6 Hz throughout.

(a) Reflected sound stays in air: lambda = v_air/f = 340/(1.0x10^6) = 3.4 x 10^-4 m (0.34 mm).

(b) Transmitted sound enters water: lambda = v_water/f = 1486/(1.0x10^6) = 1.486 x 10^-3 m (about 1.49 mm).

✦ The transmitted wavelength in water is over four times longer than in air, since frequency is fixed by the source and wavelength must scale directly with whatever speed the wave has in its current medium.
14.72 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.7
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km/s? The operating frequency of the scanner is 4.2 MHz.
Hint. Convert both given values to consistent SI units first, then apply lambda = v/f directly.
Step 1 — Convert units. v = 1.7 km/s = 1700 m/s. f = 4.2 MHz = 4.2 x 10^6 Hz.

Step 2 — Apply lambda = v/f. lambda = 1700/(4.2x10^6) = 4.05 x 10^-4 m.

✦ The wavelength inside the tissue is about 0.4 mm, small enough that ultrasonic imaging can resolve fine structures — this is exactly why scanners use megahertz frequencies rather than audible ones, which would have wavelengths far too large to image small tumours.
14.84 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.8
A transverse harmonic wave on a string is described by y(x,t) = 3.0 sin(36t + 0.018x + pi/4), where x and y are in cm and t in s. The positive direction of x is from left to right. (a) Is this a travelling wave or a stationary wave? If travelling, what are the speed and direction of its propagation? (b) What are its amplitude and frequency? (c) What is the initial phase at the origin? (d) What is the least distance between two successive crests in the wave?
Hint. Compare directly to the standard form a sin(kx +/- omega t + phi) — the sign between the kx and omega t terms fixes the direction, since a plus sign there means the wave travels in the negative x-direction.
This is a travelling wave, since kx and omega t appear together in one combined phase, not separately.

(a) Comparing to a sin(kx + omega t + phi): the + sign between 0.018x and 36t means this wave travels in the NEGATIVE x-direction. Speed v = omega/k = 36/0.018 = 2000 cm/s = 20 m/s.

(b) Amplitude a = 3.0 cm. omega = 36 rad/s, so frequency nu = omega/(2 pi) = 36/6.2832 = 5.73 Hz.

(c) At x=0, t=0, the phase is simply the constant term: pi/4.

(d) Least distance between successive crests = wavelength lambda = 2 pi/k = 2 pi/0.018 = 349 cm (about 3.49 m).

✦ The wave moves right-to-left at 20 m/s with wavelength 3.49 m, since a + sign linking kx and omega t is precisely what marks a wave as travelling in the negative x-direction rather than the positive one.
14.93 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.9
For the wave described in Exercise 14.8, describe the displacement (y) versus (t) graphs for x = 0, 2, and 4 cm. What are the shapes of these graphs? In which aspect does the oscillatory motion at one point differ from another: amplitude, frequency, or phase?
Hint. Since it is one single wave, every point shares the same amplitude and the same angular frequency omega — only the phase constant picks up an extra x-dependent contribution at each location.
At each fixed x, y(t) = 3.0 sin(36t + 0.018x + pi/4) is a sinusoidal curve in t, with the SAME amplitude (3.0 cm) and the SAME angular frequency (36 rad/s) at every location — only the constant offset inside the sine, 0.018x, differs from point to point.

At x=0: extra phase = 0. At x=2 cm: extra phase = 0.018 x 2 = 0.036 rad. At x=4 cm: extra phase = 0.018 x 4 = 0.072 rad.

✦ All three y-t graphs are identical sine curves of the same amplitude and period, merely shifted slightly in time relative to each other, since each point simply repeats the same motion the point before it made, just a little later. The oscillatory motion differs from point to point ONLY in phase, never in amplitude or frequency, which is exactly what it means for a single travelling wave to pass through every point of the medium in turn.
14.104 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.10
For the travelling harmonic wave y(x,t) = 2.0 cos[2 pi (10t - 0.0080x + 0.35)], where x and y are in cm and t in s, calculate the phase difference between the oscillatory motion of two points separated by a distance of (a) 4 m, (b) 0.5 m, (c) lambda/2, (d) 3 lambda/4.
Hint. First find the wavelength from the coefficient of x inside the cosine, then use phase difference = (2 pi/lambda) times the separation for each part.
Step 1 — Find the wavelength. The coefficient of x is 2 pi x 0.0080 = k, so lambda = 1/0.0080 = 125 cm = 1.25 m.

Step 2 — Apply phase difference = 2 pi (delta-x / lambda).

(a) delta-x = 4 m = 400 cm: phase diff = 2 pi (400/125) = 2 pi (3.2) = 6.4 pi rad (about 20.1 rad).

(b) delta-x = 0.5 m = 50 cm: phase diff = 2 pi (50/125) = 2 pi (0.4) = 0.8 pi rad (about 2.51 rad).

(c) delta-x = lambda/2: phase diff = 2 pi (1/2) = pi rad, true by definition regardless of the actual wavelength value.

(d) delta-x = 3 lambda/4: phase diff = 2 pi (3/4) = 1.5 pi rad (about 4.71 rad).

✦ Parts (c) and (d) do not even need the numeric wavelength, since expressing the separation AS a fraction of lambda directly gives the phase difference as that same fraction of 2 pi — a useful shortcut whenever separation is given in terms of lambda rather than an absolute distance.
14.114 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.11
The transverse displacement of a string (clamped at both ends) is given by y(x,t) = 0.06 sin(2 pi x/3) cos(120 pi t), where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 x 10^-2 kg. (a) Does the function represent a travelling wave or a stationary wave? (b) Interpret the wave as a superposition of two waves travelling in opposite directions — what is the wavelength, frequency, and speed of each? (c) Determine the tension in the string.
Hint. x and t appear as separate factors, not combined — that alone identifies the wave type. Compare directly to the standard form 2a sin(kx) cos(omega t) to read off amplitude, k, and omega.
(a) x and t appear as SEPARATE factors, sin(2 pi x/3) and cos(120 pi t), not combined into one phase — this is a STATIONARY (standing) wave.

(b) Comparing to y = 2a sin(kx) cos(omega t): 2a = 0.06, so a = 0.03 m for each component wave. k = 2 pi/3 per m, so lambda = 2 pi/k = 3 m. omega = 120 pi rad/s, so nu = omega/(2 pi) = 60 Hz. Speed v = lambda x nu = 3 x 60 = 180 m/s.

(c) Linear mass density mu = mass/length = (3.0x10^-2)/1.5 = 0.02 kg/m. From v = sqrt(T/mu): T = v^2 x mu = 180^2 x 0.02 = 32400 x 0.02 = 648 N.

✦ Each of the two oppositely-travelling component waves has amplitude 0.03 m, wavelength 3 m, frequency 60 Hz, and speed 180 m/s, since a standing wave is always the sum of two identical travelling waves moving in opposite directions — the string's tension, 648 N, follows from that shared speed.
14.124 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.12
(i) For the wave on the string described in Exercise 14.11, do all points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?
Hint. In a standing wave, every point shares the same angular frequency, but amplitude clearly depends on position (from the sin(kx) factor). Phase needs care: points on either side of a node move exactly opposite to each other.
(i)(a) Frequency: YES, all points oscillate with the same angular frequency, 120 pi rad/s (60 Hz) — this is the defining feature of a standing wave, since cos(omega t) has no x-dependence at all.

(i)(b) Phase: NOT all points share the same phase overall. Points WITHIN the same segment (between two adjacent nodes) oscillate in phase with each other, but points in adjacent segments, separated by a node, oscillate exactly pi out of phase (moving in opposite directions at the same instant), since sin(kx) changes sign across a node.

(i)(c) Amplitude: NO — amplitude 0.06 sin(2 pi x/3) depends explicitly on position x, ranging from zero at the nodes to a maximum of 0.06 m at the antinodes.

(ii) At x = 0.375 m: amplitude = 0.06 sin(2 pi x 0.375/3) = 0.06 sin(0.25 pi) = 0.06 x 0.7071 = 0.0424 m.

✦ The point 0.375 m from one end oscillates with amplitude about 4.24 cm, since plugging its specific x-value into the position-dependent amplitude function is what actually determines how far any given point on a standing wave swings, unlike a travelling wave where every point shares one common amplitude.
14.134 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.13
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 = 2cos(3x)sin(10t), (b) y = 2 sqrt(x - vt), (c) y = 3sin(5x - 0.5t) + 4cos(5x - 0.5t), (d) y = cos(x)sin(t) + cos(2x)sin(2t).
Hint. A travelling wave has x and t appearing ONLY through one combined variable like (x - vt) throughout the whole expression. A stationary wave is a product of a function of x alone times a function of t alone. Check each option against both definitions directly.
(a) y = 2cos(3x)sin(10t) is a clean product of a function of x alone and a function of t alone — a STATIONARY wave.

(b) y = 2 sqrt(x - vt) depends on x and t only through the single combination (x - vt) throughout — a TRAVELLING wave (a non-periodic pulse shape moving rigidly with speed v).

(c) y = 3sin(5x - 0.5t) + 4cos(5x - 0.5t): both terms depend on x and t only through the same combination (5x - 0.5t), so their sum is also purely a function of that one combination (it simplifies to a single sinusoid via A sin(theta) + B cos(theta) = R sin(theta + delta)) — a TRAVELLING wave, with speed 0.5/5 = 0.1 (in the given units).

(d) y = cos(x)sin(t) + cos(2x)sin(2t) is a SUM of two individually stationary-shaped terms (each itself a product of a function of x alone and a function of t alone), but with different k and omega in each term, so the composite does not reduce to either a single product form or a single combined-variable form — this is a SUPERPOSITION OF TWO STATIONARY WAVES, not a travelling wave.

✦ (a) is stationary, (b) and (c) are travelling waves (even though (b) is not periodic), and (d) is a superposition of two different stationary waves rather than fitting cleanly into either single category — checking whether x and t combine into ONE variable or stay as separate factors is the test that decides each case.
14.143 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.14
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 x 10^-2 kg and its linear mass density is 4.0 x 10^-2 kg/m. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?
Hint. Find the wire's length from its total mass and linear mass density first, then use the fundamental-mode formula to get the speed, and finally v = sqrt(T/mu) to get the tension.
Step 1 — Length of the wire. L = mass/mu = (3.5x10^-2)/(4.0x10^-2) = 0.875 m.

Step 2 — Speed from the fundamental frequency. nu_1 = v/(2L), so v = 2L x nu_1 = 2 x 0.875 x 45 = 78.75 m/s.

Step 3 — Tension. v = sqrt(T/mu), so T = v^2 x mu = 78.75^2 x 0.04 = 6201.6 x 0.04 = 248 N.

✦ The wire carries transverse waves at about 78.75 m/s under a tension of about 248 N, found by first recovering the wire's length from its given mass and mass density, since the fundamental-mode formula needs L explicitly.
14.155 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.15
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. (Edge effects may be neglected.)
Hint. For a tube closed at the piston end, successive resonance lengths differ by exactly lambda/2 — use that difference to find the wavelength, then apply v = lambda nu.
Step 1 — Find the wavelength from the length difference. Successive resonances for a pipe closed at one end differ by lambda/2: 79.3 - 25.5 = 53.8 cm = lambda/2, so lambda = 107.6 cm = 1.076 m.

Step 2 — Speed of sound. v = lambda x nu = 1.076 x 340 = 365.8 m/s.

✦ The speed of sound at this experiment's temperature is about 366 m/s, found entirely from the DIFFERENCE between the two resonance lengths — a method that conveniently cancels out any end-correction offset, which is exactly why the problem specifies neglecting edge effects.
14.164 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.16
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel?
Hint. Clamping the rod at its middle forces a displacement node there, with the two free ends as antinodes — figure out what fraction of a wavelength that antinode-node-antinode pattern spans across the rod's full length.
Step 1 — Relate rod length to wavelength. With a node at the clamped middle and antinodes at both free ends, the pattern antinode-node-antinode spans exactly half a wavelength across the rod's full length L: L = lambda/2, so lambda = 2L = 2 x 1.00 = 2.00 m.

Step 2 — Speed of sound in steel. v = lambda x nu = 2.00 x 2530 = 5060 m/s.

✦ The speed of longitudinal sound in this steel rod is about 5060 m/s, consistent with steel's known high speed of sound — the key step is recognising that a node at the centre and antinodes at both free ends together span half a wavelength, not a full one.
14.174 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.17
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is 340 m/s.)
Hint. Find the closed pipe's fundamental frequency first and compare it to the source frequency — a small mismatch is expected and is explained by end correction. Then separately check the open-pipe harmonic series for any match at all.
Step 1 — Closed-pipe fundamental. nu_1 = v/(4L) = 340/(4 x 0.20) = 340/0.8 = 425 Hz.

Step 2 — Compare to the source. 430 Hz is very close to 425 Hz, so the source resonantly excites the FIRST (fundamental) harmonic of the closed pipe; the small 5 Hz discrepancy is attributable to end correction, since the antinode actually forms slightly beyond the physical open end.

Step 3 — Open-pipe case. With both ends open: nu_1 = v/(2L) = 340/0.4 = 850 Hz, with harmonics at 850, 1700, 2550 Hz, etc. Since 430 Hz does not match any of these, NO resonance would occur with both ends open.

✦ The 430 Hz source resonates with the closed pipe's fundamental mode (with the small mismatch explained by end correction) but would not resonate at all if both ends were opened, since opening the second end doubles the fundamental frequency and shifts the entire harmonic ladder away from 430 Hz.
14.183 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.18
Two sitar strings A and B playing the note 'Ga' are slightly out of tune and produce beats of frequency 6 Hz. The tension in string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Hint. Reducing tension in a string lowers its frequency. Work out whether B's frequency must be above or below A's from the direction the beat frequency changed.
Step 1 — Effect of reducing tension. Reducing A's tension lowers nu_A (since v = sqrt(T/mu) decreases, and fundamental frequency depends on v).

Step 2 — Determine which frequency is higher. The beat frequency DECREASED as nu_A decreased. This can only happen if nu_A was moving TOWARD nu_B, meaning nu_B was already below nu_A. (If nu_B were above nu_A, decreasing nu_A further would have increased the gap, not decreased it.)

Step 3 — Solve for nu_B. nu_A - nu_B = 6, so nu_B = 324 - 6 = 318 Hz.

✦ String B's frequency is 318 Hz, found by reasoning about which direction the beat frequency moved — this kind of directional reasoning, not just the beat-frequency formula itself, is what these problems are actually testing.
14.195 marksNCERT Cl-11 Physics Part II, Ch14 Exercises, Q14.19
Explain why (or how): (a) in a sound wave, a displacement node is a pressure antinode and vice versa, (b) bats can ascertain distances, directions, nature, and sizes of obstacles without any 'eyes', (c) a violin note and a sitar note may have the same frequency, yet we can distinguish between the two notes, (d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, (e) the shape of a pulse gets distorted during propagation in a dispersive medium.
Hint. Each part has a distinct one-idea explanation: (a) pressure depends on how much neighbouring layers compress together, not on displacement itself; (d) is about which elastic modulus each wave type needs; (e) is about whether wave speed depends on frequency.
(a) Pressure variation depends on the SPATIAL GRADIENT of displacement (how much neighbouring layers of air are compressed together), not on displacement itself. At a displacement node, particles on either side move maximally toward or away from that point, producing the largest compression there — a pressure antinode. At a displacement antinode, nearby particles move together with little relative compression, so pressure barely changes there — a pressure node.

(b) Bats emit ultrasonic pulses and listen for the echo. The time delay reveals distance (via the known speed of sound), comparing what each ear hears reveals direction, and the intensity, frequency shift, and pattern of the returning echo reveal information about the size and nature of the obstacle — all without needing to see it directly.

(c) Two instruments playing the same note share the same FUNDAMENTAL frequency (pitch), but each produces a different relative mix of overtones (harmonics) superposed on that fundamental. This difference in waveform, called timbre or tone quality, is what lets the ear distinguish the two instruments even at identical pitch.

(d) A transverse wave needs the medium to resist shear (sideways) deformation and spring back — a property measured by shear modulus, which only solids possess. A longitudinal wave only needs a bulk (compressive) modulus, which every state of matter has, including gases — this is why gases support only longitudinal waves while solids support both.

(e) A pulse is not a single frequency — by Fourier's theorem it is a superposition of many different sinusoidal frequency components. In a dispersive medium, wave speed depends on frequency, so each component travels at a slightly different speed and the components gradually spread apart relative to each other, changing the pulse's overall shape as it propagates. In a non-dispersive medium, all components travel at the same speed and the pulse shape stays fixed.

✦ Each explanation traces back to a single underlying idea — spatial gradients for (a), reflected-wave information for (b), harmonic content for (c), which elastic modulus is available for (d), and frequency-dependent speed for (e) — rather than needing separate unrelated facts for each part.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part II textbook, Reprint 2026-27 (keph207.pdf, 22 pages, "CHAPTER FOURTEEN"), cross-checked against the official CBSE curriculum 2026-27, Subject Code 042, Unit X. One end-of-chapter Exercises set (19 questions, 14.1-14.19). CBSE's syllabus line for this chapter (transverse/longitudinal waves, speed of a travelling wave, displacement relation, superposition, reflection, standing waves in strings and organ pipes, harmonics, beats) maps directly onto NCERT's sections 14.2-14.7 in the same order; no content gap found. The Doppler effect is absent from both the chapter body and the CBSE syllabus line for this chapter — not an oversight, just not examinable content here.. Questions are referenced from the NCERT textbook for identification.

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