Class 11 Physics · Chapter 14
Short answer:
Quick answer: Class 11 Physics Chapter 14 'Waves' mechanical waves — transverse aur longitudinal — unki displacement relation y = a sin(kx − ωt), wave speed (string: v = √(T/μ); sound in gas: v = √(γP/ρ)), principle of superposition, standing waves aur harmonics (string, open/closed organ pipe), beats, aur Doppler effect cover karta hai. Ye 14-chapter rationalised syllabus (2026-27 session) ka aakhri chapter hai — purana 'Physical World' Chapter 1 hata diya gaya hai, isliye numbering old PDFs se match nahi karegi (fresh class 11 physics all chapters pdf 2026-27 hi use karo). Exact class 11 physics chapter wise weightage CBSE board exam me har saal thoda badalta hai — is spec me sirf verified concept-depth aur registry-grade syllabus-structure di gayi hai, koi fabricated number nahi.
Chapter 14 'Waves' Class 11 Physics ka aakhri chapter hai — is rationalised 2026-27 syllabus me, jisme ab total 14 chapters hain aur purana 'Physical World' wala Chapter 1 hi drop kar diya gaya hai. Ye chapter seedha peeche wale Chapter 13 'Oscillations' ke SHM concept ko utha ke space me travel karwata hai — ek jagah ka dolna (oscillation) yaha ek pura medium ka dolna (wave) ban jaata hai. Isi chapter se sound, music instrument ki timbre, ultrasound scanning, aur radar/Doppler jaisi real-duniya cheezein samajh me aati hain, aur Class 12 ke optics + AC (alternating current) chapters ki wave-thinking ki neev bhi yahi hai. Agar tum class 11 physics ncert solutions chapter-by-chapter dhundh rahe ho, toh Waves ko halka mat lo — string pe standing waves, organ-pipe harmonics, beats, aur Doppler effect — ye sab har saal numericals ka strong source hote hain.
Chapter 14 Summary — 5 Minute Revision
Waves chapter transverse aur longitudinal waves ke basic classification se shuru hokar, ek progressive wave ki displacement relation y(x,t) = a sin(kx − ωt + φ) tak leke jaata hai, jisme k (angular wave number) aur ω (angular frequency) wave ki spatial aur temporal periodicity ko describe karte hain. String pe transverse wave ki speed v = √(T/μ) aur gas me sound ki speed v = √(γP/ρ) — dono hi tension/elasticity aur inertia (density) ke beech ka trade-off dikhate hain.
Principle of superposition batata hai ki do ya zyada waves ek hi medium me independently add ho jaati hain (y = y₁+y₂+...) — isi se do practically important phenomena nikalte hain: standing (stationary) waves, jab equal amplitude/frequency ki do waves opposite directions me milti hain (string ke fixed ends, ya organ pipe ke reflection se), aur beats, jab thodi different frequency ki do waves same direction me superpose hoti hain, jisse loud-soft-loud periodic variation sunayi deti hai (beat frequency = |ν₁−ν₂|).
Standing waves se harmonics/overtones ka concept aata hai — string fixed at both ends aur open organ pipe sabhi harmonics (νₙ = nv/2L) support karte hain, jabki ek end band organ pipe sirf odd harmonics (νₙ = (2n−1)v/4L). Chapter ka aakhri bada topic Doppler effect hai — jab source ya observer (ya dono) medium ke relative move karte hain, toh observed frequency source-frequency se badal jaati hai; is chapter me is formula ka derivation rationalised syllabus me simplify kiya gaya hai, lekin core formulas (source/observer approaching-receding) exam-relevant rehte hain.
In-Text Questions — Solutions
Example 14.1 A wave travelling along a string is described by y(x,t) = 0.005 sin(80.0x − 3.0t), numerical constants in SI units. Find (a) amplitude, (b) wavelength, (c) period, (d) speed.
Comparing with y = a sin(kx − ωt): a = 0.005 m, k = 80.0 rad/m, ω = 3.0 rad/s
(a) Amplitude a = 5 mm
(b) λ = 2π/k = 2π/80.0 ≈ 7.85 cm
(c) T = 2π/ω = 2π/3.0 ≈ 2.09 s
(d) v = ω/k = 3.0/80.0 = 0.0375 m/s
Example 14.2 A steel wire 0.72 m long has a mass of 5.0×10⁻³ kg. If the wire is under a tension of 60 N, find the speed of transverse waves on it.
μ = m/L = 5.0×10⁻³/0.72 = 6.94×10⁻³ kg/m
v = √(T/μ) = √(60/6.94×10⁻³) = √8646 ≈ 93.0 m/s
Example 14.3 Two identical waves of amplitude a and same frequency travel in opposite directions on a string. Show the resultant is y = 2a sin(kx)cos(ωt), and find the node positions if λ = 2 m.
y₁ = a sin(kx − ωt), y₂ = a sin(kx + ωt)
By superposition, y = y₁ + y₂ = a[sin(kx − ωt) + sin(kx + ωt)]
Using sinA + sinB = 2 sin((A+B)/2) cos((A−B)/2):
y = 2a sin(kx) cos(ωt)
Nodes occur where sin(kx) = 0 ⟹ kx = nπ ⟹ x = nλ/2, n = 0,1,2,...
With λ = 2 m, nodes are at x = 0, 1, 2, 3, ... m (spaced 1 m apart).
Example 14.4 A wire of linear mass density 4×10⁻² kg/m, fixed between rigid supports 1.0 m apart, vibrates in its fundamental mode at 100 Hz. Find the tension and the frequency of the first overtone.
Fundamental: ν₁ = v/2L ⟹ v = 2Lν₁ = 2×1.0×100 = 200 m/s
T = v²μ = (200)² × 4×10⁻² = 1600 N
First overtone = 2nd harmonic: ν₂ = 2ν₁ = 200 Hz
Example 14.5 A train's whistle emits 500 Hz while approaching a stationary observer at 20 m/s. Speed of sound = 340 m/s. Find the frequency heard.
Source moving toward stationary observer:
ν' = ν·v/(v − v_s) = 500 × 340/(340 − 20) = 500 × 340/320 ≈ 531 Hz
Example 14.6 (Beats) Tuning fork A (256 Hz) and fork B together give 4 beats/s. Loading fork B with a little wax reduces the beat frequency to 2 beats/s. Find the frequency of B.
|ν_A − ν_B| = 4 Hz, ν_A = 256 Hz ⟹ ν_B = 260 Hz or 252 Hz.
Loading with wax lowers ν_B (adds mass).
If ν_B = 260 Hz: lowering it moves it closer to 256 → beats decrease — consistent.
If ν_B = 252 Hz: lowering it moves it further from 256 → beats would increase — inconsistent.
Hence ν_B = 260 Hz.
Poore Class 11 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q19)
14.1 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?
Linear mass density:
μ = m/L = 2.50/20.0 = 0.125 kg/m
Speed of transverse wave on string:
v = √(T/μ) = √(200/0.125) = √1600 = 40 m/s
Time to travel length L:
t = L/v = 20.0/40 = 0.5 s
14.2 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/s²)
Time for stone to fall (u = 0):
h = ½gt₁² ⟹ t₁ = √(2h/g) = √(2×300/9.8) = √61.2 = 7.82 s
Time for sound of splash to travel back up:
t₂ = h/v = 300/340 = 0.88 s
Total time from release to hearing splash:
t = t₁ + t₂ = 7.82 + 0.88 = 8.71 s
14.3 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 equals the speed of sound in dry air at 20 °C = 343 m/s?
μ = m/L = 2.10/12.0 = 0.175 kg/m
From v = √(T/μ):
T = v²μ = (343)² × 0.175 = 117649 × 0.175 = ≈ 2.06 × 10⁴ N
14.4 Use the formula v = √(γP/ρ) to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.
(a) For an ideal gas, P = ρRT/M, so P/ρ = RT/M — this depends only on temperature T, not on P separately. Since v = √(γP/ρ) = √(γRT/M), at constant T, v stays constant even if P changes (P and ρ change together, keeping P/ρ fixed).
(b) Since v = √(γRT/M), v ∝ √T. As temperature rises, average molecular speed rises, so v increases.
(c) Moist air contains water vapour (molar mass 18) which is lighter than the average molar mass of dry air (≈28.8). This lowers the effective mean molar mass M of humid air. Since v ∝ 1/√M, a lower M means a higher v — hence sound travels faster in humid air.
14.5 A travelling wave in one dimension is represented by y = f(x − vt). Is the converse true — does every function of x and t of this form represent a travelling wave? Examine: (a) (x − vt)², (b) log[(x + vt)/x₀], (c) 1/(x + vt).
No, the converse is not automatically true. A function must remain finite everywhere (for all x, t) to physically represent a travelling wave, even if it has the mathematical form f(x ∓ vt).
(a) (x − vt)² → as x → ∞, this → ∞. Unbounded, so it does not represent a physically valid travelling wave.
(b) log[(x+vt)/x₀] → as (x+vt) → 0, this → −∞. Unbounded, so it does not represent a travelling wave.
(c) 1/(x+vt) → as (x+vt) → 0, this → ∞. Unbounded, so it does not represent a travelling wave.
None of the three qualify — each blows up (diverges) somewhere, whereas a genuine wave function must stay finite everywhere.
14.6 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 = 340 m/s, in water = 1486 m/s)
Frequency stays the same on reflection and on transmission across a boundary — only speed and wavelength change.
(a) Reflected sound (stays in air, v = 340 m/s):
λ = v/ν = 340/(1000×10³) = 3.4×10⁻⁴ m = 0.34 mm
(b) Transmitted sound (enters water, v = 1486 m/s):
λ = 1486/(1000×10³) = 1.486×10⁻³ m ≈ 1.49 mm
14.7 A hospital uses an ultrasonic scanner to locate tumours in tissue where the speed of sound is 1.7 km/s. The operating frequency of the scanner is 4.2 MHz. What is the wavelength of sound in the tissue?
v = 1.7 × 10³ m/s, ν = 4.2 × 10⁶ Hz
λ = v/ν = 1700/(4.2×10⁶) = 4.05×10⁻⁴ m ≈ 0.40 mm
14.8 A transverse harmonic wave on a string is described by y(x,t) = 3.0 sin(36t + 0.018x + π/4), x, y in cm and t in s. (a) Is this travelling or stationary? If travelling, find speed and direction. (b) Amplitude and frequency. (c) Initial phase at the origin. (d) Least distance between two successive crests.
Compare with y = a sin(ωt + kx + φ): a = 3.0 cm, ω = 36 rad/s, k = 0.018 rad/cm, φ = π/4.
(a) It is a travelling wave (single sinusoidal argument linear in both x and t, not a product of separate x-function and t-function). Since kx and ωt have the same sign, the wave travels in the −x direction.
v = ω/k = 36/0.018 = 2000 cm/s = 20 m/s
(b) Amplitude = 3.0 cm; frequency ν = ω/2π = 36/2π ≈ 5.73 Hz
(c) Initial phase φ = π/4 rad = 45°
(d) Distance between successive crests = wavelength:
λ = 2π/k = 2π/0.018 ≈ 349 cm ≈ 3.5 m
14.9 For the wave in Exercise 14.8, sketch y–t graphs at x = 0, 2, 4 cm. What shape do these graphs have? In a travelling wave, do amplitude, frequency, or phase differ from one point to another?
At any fixed x, y(t) = 3.0 sin(36t + 0.018x + π/4) is a pure sine function of t — so each graph is a sinusoidal curve of amplitude 3 cm and period T = 2π/36 ≈ 0.175 s.
All three graphs (x = 0, 2, 4 cm) are identical sine curves, only shifted along the time-axis relative to each other, by phase differences Δφ = kΔx = 0.018×2 = 0.036 rad (x=0 to x=2) and 0.018×4 = 0.072 rad (x=0 to x=4).
Conclusion: In a travelling wave, every point oscillates with the same amplitude and same frequency — only the phase differs from point to point.
14.10 For the travelling wave y(x,t) = 2.0 cos 2π(10t − 0.0080x + 0.35), x,y in cm, t in s, find the phase difference between two points separated by (a) 4 m, (b) 0.5 m, (c) λ/2, (d) 3λ/4.
Comparing with y = a cos 2π(t/T − x/λ + φ₀): ν = 10 Hz, and 1/λ = 0.0080 cm⁻¹ ⟹ λ = 125 cm = 1.25 m.
Phase difference for separation Δx: Δφ = (2π/λ)Δx
(a) Δx = 4 m = 3.2λ → Δφ = 3.2 × 2π ≈ 20.1 rad
(b) Δx = 0.5 m = 0.4λ → Δφ = 0.4 × 2π ≈ 2.51 rad
(c) Δx = λ/2 → Δφ = π ≈ 3.14 rad (independent of actual λ value)
(d) Δx = 3λ/4 → Δφ = 3π/2 ≈ 4.71 rad
14.11 The transverse displacement of a string (clamped at both ends) is y(x,t) = 0.06 sin(2πx/3) cos(120πt), x,y in m, t in s. Length = 1.5 m, mass = 3.0×10⁻² kg. (a) Travelling or stationary wave? (b) Interpret as superposition of two opposite travelling waves — find λ, ν, v of each. (c) Find the tension in the string.
(a) y is a product of a function of x alone and a function of t alone → this is a stationary (standing) wave, not a travelling one.
(b) Using sinA·cosB = ½[sin(A+B) + sin(A−B)]:
y = 0.03 sin(2πx/3 − 120πt) + 0.03 sin(2πx/3 + 120πt)
Each component wave has amplitude 0.03 m, k = 2π/3 m⁻¹ ⟹ λ = 3.0 m; ω = 120π ⟹ ν = 60 Hz; speed:
v = ω/k = 120π/(2π/3) = 180 m/s (travelling in opposite directions)
(c) μ = m/L = 3.0×10⁻²/1.5 = 0.02 kg/m
T = v²μ = (180)² × 0.02 = 648 N
14.12 (i) For the wave in Exercise 14.11, do all points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain. (ii) What is the amplitude of vibration at x = 0.375 m?
(i)(a) Frequency: Yes — every point has the common factor cos(120πt), so all oscillate at the same frequency, 60 Hz.
(b) Phase: All points between two consecutive nodes vibrate in phase with each other; points in adjacent loops (segments), separated by a node, vibrate in opposite phase (180° apart), because sin(2πx/3) changes sign across a node.
(c) Amplitude: No — amplitude = |0.06 sin(2πx/3)| varies with x: it is zero at nodes and maximum (0.06 m) at antinodes.
(ii) At x = 0.375 m:
Amplitude = 0.06 sin(2π×0.375/3) = 0.06 sin(45°) = 0.06 × 0.707 ≈ 0.042 m (4.2 cm)
14.13 State which of the following represent (i) a travelling wave, (ii) a stationary wave, (iii) neither: (a) y = 2cos(3x)sin(10t), (b) y = 2√(x−vt), (c) y = 3sin(5x−0.5t) + 4cos(5x−0.5t), (d) y = cos x sin t + cos 2x sin 2t.
(a) y = 2cos(3x)·sin(10t) is a product of a function of x and a function of t → stationary wave.
(b) y = 2√(x−vt) is purely a function of (x−vt) → travelling wave (moving in +x direction).
(c) Using R sinθ + R'cosθ form: 3sin(5x−0.5t) + 4cos(5x−0.5t) = 5 sin(5x − 0.5t + φ), where tanφ = 4/3. This is a single sinusoid in argument (5x − 0.5t) → travelling wave.
(d) This is the sum of two different stationary waves (cos x sin t has k=1, ω=1; cos 2x sin 2t has k=2, ω=2) — since they have different λ and ν, the sum is neither a single travelling wave nor a single stationary wave.
14.14 A wire stretched between two rigid supports vibrates in its fundamental mode with frequency 45 Hz. Mass of wire = 3.5×10⁻² kg, linear mass density = 4×10⁻² kg/m. Find (a) the speed of the transverse wave, (b) the tension in the string.
Length: L = m/μ = 3.5×10⁻²/4×10⁻² = 0.875 m
(a) Fundamental mode: ν₁ = v/2L
v = 2Lν₁ = 2 × 0.875 × 45 = 78.75 m/s
(b) T = v²μ = (78.75)² × 4×10⁻² = 6201.6 × 0.04 ≈ 248 N
14.15 A metre-long tube, open at one end with a movable piston at the other end, resonates with a 340 Hz tuning fork when the air-column length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air.
This is effectively a pipe closed at the piston end. Successive resonance lengths are separated by λ/2:
λ/2 = 79.3 − 25.5 = 53.8 cm ⟹ λ = 107.6 cm = 1.076 m
v = νλ = 340 × 1.076 ≈ 366 m/s
(Using the difference of two resonance lengths automatically cancels the end-correction.)
14.16 A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibration of the rod is 2.53 kHz. What is the speed of sound in steel?
Clamping at the middle fixes a node there, with antinodes at both free ends — this fundamental mode pattern spans half a wavelength over the full rod length:
L = λ/2 ⟹ λ = 2L = 2 × 1.00 = 2.00 m
v = νλ = 2.53×10³ × 2.00 = 5.06 × 10³ m/s
14.17 A pipe 20 cm long is closed at one end. Which harmonic mode is resonantly excited by a 430 Hz source? Would the same source resonate with the pipe if it were open at both ends? (speed of sound = 340 m/s)
Closed pipe, L = 0.20 m: resonant frequencies νₙ = (2n−1)v/4L, n=1,2,3,... (only odd harmonics).
ν₁ (fundamental) = v/4L = 340/(4×0.20) = 425 Hz
The source at 430 Hz is very close to 425 Hz, so it resonantly excites the first harmonic (fundamental mode, n = 1) of the closed pipe.
Open pipe of same length: νₙ' = nv/2L = n × 340/0.40 = n × 850 Hz. The fundamental itself is 850 Hz, far from 430 Hz, and no higher harmonic (850, 1700, ...) is close either — so no resonance would occur with an open pipe.
14.18 Two sitar strings A and B produce 6 beats/s. When the tension in A is slightly reduced, the beat frequency drops to 3 beats/s. If the frequency of A is 324 Hz, find the frequency of B.
Beat frequency = |ν_A − ν_B| = 6 Hz, with ν_A = 324 Hz ⟹ ν_B = 330 Hz or ν_B = 318 Hz.
Reducing tension in A decreases ν_A (since ν ∝ √T).
If ν_B = 330 Hz: as ν_A falls below 324, the gap |ν_A − 330| would increase → beats would increase — contradicts the observed decrease.
If ν_B = 318 Hz: as ν_A falls from 324 toward 318, the gap |ν_A − 318| decreases → beats decrease — matches the observation.
Therefore, ν_B = 318 Hz.
14.19 Explain: (a) why a displacement node is a pressure antinode and vice versa; (b) how bats judge distance/direction/size of obstacles without eyes; (c) why a violin note and a sitar note of the same pitch and intensity still sound different; (d) why solids support both longitudinal and transverse waves but gases support only longitudinal waves; (e) why a pulse gets distorted while travelling through a dispersive medium.
(a) At a displacement node, particles on either side move toward or away from each other, creating maximum compression/rarefaction there — so it is a pressure antinode. At a displacement antinode, neighbouring particles move together in phase with no relative compression, so pressure variation is minimum there — a pressure node.
(b) Bats emit high-frequency ultrasonic waves that reflect off obstacles; by timing the echo's return and analysing its intensity/Doppler shift/direction, bats judge distance, direction, and roughly the size/nature of the obstacle — natural echolocation (sonar).
(c) Though fundamental frequency (pitch) and loudness are equal, each instrument produces a different mix of overtones/harmonics superposed on the fundamental. This different harmonic content gives each instrument a distinct waveform shape (timbre/quality), letting the ear tell them apart.
(d) Transverse waves need a restoring shear force. Solids have shear elasticity (rigidity), so they support both longitudinal and transverse waves. Gases have no shear elasticity — they resist only volume compression (bulk elasticity) — so gases can propagate only longitudinal waves.
(e) In a dispersive medium, wave speed depends on frequency. A pulse is a superposition of many frequency components; since each travels at a different speed, their relative phases change as they propagate, so the pulse's shape progressively distorts (spreads out).
Important Equations — Ek Nazar Me
Note: v = wave speed, ν = frequency, λ = wavelength, T = tension, μ = linear mass density, γ = ratio of specific heats, P = pressure, ρ = density, a = amplitude, k = 2π/λ, ω = 2πν, L = length, n = harmonic number, v_s/v_o = source/observer speed.
| Concept | Formula |
|---|---|
| Wave speed relation | v = νλ |
| Transverse wave on stretched string | v = √(T/μ), where μ = m/L |
| Speed of sound in a gas (Laplace) | v = √(γP/ρ) |
| General progressive wave | y(x,t) = a sin(kx − ωt + φ) |
| Angular wave number | k = 2π/λ |
| Angular frequency | ω = 2πν = 2π/T |
| Principle of superposition | y = y₁ + y₂ (+ y₃ + ...) |
| Stationary wave (string, both ends fixed) | y = 2a sin(kx) cos(ωt) |
| Harmonics — string fixed at both ends | νₙ = nv/2L, n = 1, 2, 3, ... (all harmonics) |
| Closed organ pipe (one end closed) | νₙ = (2n−1)v/4L (only odd harmonics) |
| Open organ pipe (both ends open) | νₙ = nv/2L (all harmonics) |
| Beat frequency | ν_beat = |ν₁ − ν₂| |
| Doppler — source approaching stationary observer | ν' = ν·v/(v − v_s) |
| Doppler — source receding | ν' = ν·v/(v + v_s) |
| Doppler — observer approaching stationary source | ν' = ν·(v + v_o)/v |
| Doppler — observer receding | ν' = ν·(v − v_o)/v |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Wave-equation ke sign ko galat padhna — y = a sin(kx − ωt) −x direction ki wave hoti hai jab kx aur ωt ka sign same ho (jaise kx+ωt), opposite sign ho toh +x direction; students direction ulta bata dete hain.
- String (both ends fixed) ke harmonics me spacing λ/2 hoti hai, jabki closed organ pipe me antara λ/4 (fundamental) — dono ko confuse karke wrong length/frequency nikal lete hain.
- Closed organ pipe (ek end band) sirf ODD harmonics deta hai (νₙ = (2n−1)v/4L) — students isme bhi galti se saare (odd+even) harmonics allow kar dete hain jaisa open pipe me hota hai.
- Doppler formula me numerator-denominator ulta kar dena — source approach karne pe apparent frequency badhni chahiye, isliye denominator me (v − v_s) aata hai na ki (v + v_s); ye sabse common calculation mistake hai.
- Linear mass density μ nikalte waqt poore string ka total mass use kar lena instead of mass/length (μ = m/L) — is wajah se tension ya speed ka jawab off by a large factor aa jaata hai.
- Beat frequency nikalte waqt angular frequency (ω₁ − ω₂) ka difference le lena instead of ordinary frequency (ν₁ − ν₂), ya phir absolute value lagana bhool jaana jab ν₂ > ν₁ ho jaaye.
Board-Style Important Questions
- 1 mark: A pipe closed at one end resonates with which set of harmonics — only odd, only even, or all? (Ans: only odd harmonics, νₙ = (2n−1)v/4L)
- 2 marks: Define beats and write the formula for beat frequency, using an example of two tuning forks of slightly different frequency.
- 3 marks: Derive the expression v = √(T/μ) for the speed of a transverse wave on a stretched string using dimensional analysis, and state the assumptions made.
- 3 marks: Two waves of equal amplitude and frequency travel in opposite directions along a stretched string. Show that their superposition produces a stationary wave y = 2a sin(kx)cos(ωt), and locate the positions of nodes and antinodes.
- 5 marks: A string of given length, mass and tension is fixed at both ends. Find the fundamental frequency and the frequencies of the first two overtones (numerical, values as given in the question).
- 5 marks: State the Doppler effect for sound. Derive the apparent frequency heard by a stationary observer when a source moves toward and then away from the observer, and solve for the frequency shift given source speed and speed of sound.
Aksar Poochhe Jaane Wale Sawaal
NCERT Class 11 Physics ke saare chapters ek PDF me 2026-27 session ke liye kahan milenge?
Rationalised (2026-27) Class 11 Physics me total 14 chapters hain — 'Physical World' wala purana Chapter 1 hata diya gaya hai, ab book seedha 'Units and Measurements' se shuru hoti hai. Class 11 physics all chapters pdf 2026-27 ke liye sabse reliable source official ncert.nic.in textbook section hai — waha se hi latest, rationalised edition download karo, kyunki third-party sites purani (pre-2023) numbering wali PDF bhi carry karti hain jo chapter-number mismatch create karti hai.
Waves chapter me Doppler effect ka formula kya hai, aur kya ye 2026-27 syllabus se pura hata diya gaya hai?
Nahi, poora topic delete nahi hua — sirf derivation ka depth reduce hua hai rationalisation me. Waves class 11 physics Doppler effect formula abhi bhi syllabus me hai:
Source approaching stationary observer: ν' = ν·v/(v − v_s)
Source receding: ν' = ν·v/(v + v_s)
Full multi-case derivation (wind, dono simultaneously moving, jaisi harder numericals) trim hui hai, par basic formula-based concept-check exam me aa sakta hai.
Class 11 Physics ka chapter-wise weightage CBSE board exam me kitna hota hai?
Class 11 physics chapter wise weightage CBSE ka koi single fixed number nahi diya ja sakta yahan honestly — official weightage/marking scheme CBSE har academic year apne sample paper aur marking scheme document me release karta hai, aur woh thoda change hota rehta hai. Is spec me hum sirf verified-registry-grade syllabus structure de rahe hain (14 chapters, kya trim hua); exact marks-split ke liye latest CBSE sample paper hi authoritative source hai — kisi bhi number ko yahan estimate/guess nahi kiya gaya.
Waves padhne se pehle konse chapters strong hone chahiye — jaise Motion in a Straight Line ya Oscillations?
Waves seedha Oscillations (Chapter 13) ke SHM concepts pe khada hai (y = a sin(ωt) wahi se aata hai), isliye pehle oscillations class 11 physics important questions revise kar lo. Neeche se foundation bhi zaroori hai — velocity/acceleration ka basic sense class 11 physics chapter 2 motion in a straight line ncert solutions se aata hai, jo graphs aur relative-motion intuition banata hai jo wave-speed problems me kaam aata hai.
Class 11 Physics ka 2026-27 deleted/rationalised syllabus me exactly kya-kya hata hai — sirf Waves ya poora syllabus?
Class 11 physics deleted syllabus 2026-27 sirf Waves tak limited nahi hai — poore syllabus me spread hai (registry-grade confirmation, official ncert.nic.in PDF se direct verify nahi ho paya is run me):
- Ch.1 'Physical World' — poora chapter hi drop, ab Units and Measurements se start
- Units and Measurements — measurement accuracy/precision ka kuch detail trim
- Motion in a Plane/Straight Line — relative-velocity ka kuch content trim
- Gravitation — geostationary satellite portion trim
- Thermodynamics — heat-engine/refrigerator sections simplify
- System of Particles and Rotational Motion — rolling-motion derivation simplify
- Waves — Doppler Effect derivation complexity reduce
Numerically precise cheez (jaise marks-weightage) publish karne se pehle ncert.nic.in ki actual PDF se cross-check recommend kiya jaata hai.
Waves ke alawa Class 11 Physics me aur kaunse chapters me heavy numericals aate hain jinki practice zaroori hai?
Waves jaisa hi numerical-heavy treatment in chapters me bhi milta hai — practice inhe bhi karo: gravitation class 11 numericals with solutions (satellite/escape velocity type), mechanical properties of fluids class 11 numericals (Bernoulli, viscosity), system of particles and rotational motion important formulas (moment of inertia, torque), kinetic theory of gases class 11 derivation (pressure expression, degrees of freedom), laws of motion class 11 ncert exemplar (friction, pulley systems), thermodynamics class 11 physics notes (first law, specific heats), motion in a plane projectile motion class 11 solutions, aur units and measurements class 11 important questions (significant figures, dimensional analysis) — ye sab mil ke Class 11 Physics ka numerical backbone banate hain, Waves sirf inka natural extension hai.
Class 11 Physics — Saare Chapters
Board exam tak sirf revision karna hai?
Class 11 Physics ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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