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अध्याय 14: तरंगें (Waves)

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CBSE • कक्षा XI • Physics • अध्याय 14

अनुमानित समय: 45 Mins

प्रगति: अध्ययनरत

तरंगें (Waves)

In Class 11 Physics, "Waves" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें

प्रस्तावना अध्याय अवधारणाएँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

🌊 Have You Ever Wondered?

How do noise-canceling headphones silence roaring jet engine airplane noise using destructive wave interference, or why does a train horn shift from high pitch to low pitch as it speeds past a station platform? Wave kinematics and acoustic physics explain wave transmission.

यह अध्याय क्यों महत्वपूर्ण है

In Class 11 Physics, "Waves" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्ययन से पूर्व (आवश्यक ज्ञान)

Oscillations from Chapter 13.
Frequency, wavelength, and speed.
Principle of superposition.

इस अध्याय के लक्ष्य

Distinguish between Transverse waves (crests/troughs, polarization) and Longitudinal waves (compressions/rarefactions).
Write equation of a Progressive Harmonic Wave: $y(x, t) = A\sin(kx - \omega t + \phi)$ and find wave speed $v = \frac{\omega}{k} = \nu\lambda$.
Apply Newton's formula for speed of sound and Laplace's correction ($v = \sqrt{\frac{\gamma P}{\rho}}$).
Apply the Principle of Superposition to analyze Standing (Stationary) Waves in stretched strings and organ pipes.
Analyze Beats: $f_{\text{beat}} = |f_1 - f_2|$.

अध्याय रूपरेखा एवं प्रगति

1 1. Traveling Waves & Wave Velocity

2 2. Speed of Sound & Laplace's Corre...

3 3. Standing Waves & Beats

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Traveling Waves & Wave Velocity

A Wave is a traveling disturbance transferring energy and momentum through a medium without net transport of matter:
• Transverse: Particle oscillations perpendicular to wave propagation (Light, waves on a string).
• Longitudinal: Particle oscillations parallel to propagation (Sound waves in air).
• Displacement Equation: $$\mathbf{y(x, t) = A\sin(kx - \omega t + \phi)} \quad \left(k = \frac{2\pi}{\lambda}, \omega = 2\pi\nu\right)$$ Wave speed: $\mathbf{v = \frac{\omega}{k} = \nu\lambda}$.

2. Speed of Sound & Laplace's Correction

Newton assumed sound propagation in air is isothermal ($v = \sqrt{P/\rho} = 280\text{ m/s}$, failing vs experimental $332\text{ m/s}$). Laplace corrected this: compressions and rarefactions happen so rapidly that heat cannot exchange (Adiabatic!): $$\mathbf{v = \sqrt{\frac{\gamma P}{\rho}}} \quad (\gamma = 1.41 \text{ for air} \implies v \approx 332\text{ m/s}!)$$

3. Standing Waves & Beats

Standing Waves (Superposition): Incident and reflected waves interfere: $y = (2A\sin kx)\cos\omega t$. Generates motionless Nodes ($x = n\lambda/2$) and maximum vibration Antinodes.
• Closed organ pipe: only odd harmonics ($1 : 3 : 5$).
• Open organ pipe: all harmonics ($1 : 2 : 3 : 4$).
Beats: Waxing and waning of sound intensity from slightly different frequencies: $$\mathbf{f_{\text{beat}} = |f_1 - f_2|}$$

Waves - Key Conceptual & Analytical Model

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1

Wave Equation: $y = A\sin(kx - \omega t)$ with propagation velocity $v = \omega/k = \nu\lambda$.

मुख्य बिंदु 2

Laplace Correction: Adiabatic bulk modulus $\gamma P$ yielding true sound velocity ($332\text{ m/s}$).

मुख्य बिंदु 3

Nodes and Antinodes: Interference boundaries of zero motion vs maximum amplitude vibration.

मुख्य बिंदु 4

Organ Pipe Harmonics: Open pipes produce all harmonics; closed pipes produce only odd frequencies.

मुख्य बिंदु 5

Beat Frequency: $f_{beat} = |f_1 - f_2|$ used in precision musical instrument acoustic tuning.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

Explain Laplace's correction to Newton's formula for the speed of sound in air.

उत्तर एवं व्याख्या देखें

उत्तर: Newton assumed sound propagation is an isothermal process ($B = P$), which yielded $v = \sqrt{P/\rho} \approx 280\text{ m/s}$, 15% lower than experimental values. Laplace recognized that compressions and rarefactions occur so rapidly that heat cannot transfer, making it an adiabatic process ($B = \gamma P$). Hence, $v = \sqrt{\frac{\gamma P}{\rho}} \approx 332.5\text{ m/s}$, matching experiment perfectly.
Sound is adiabatic (not isothermal), giving v = √(γP/ρ).

A string of mass 2.5 kg is under a tension of 200 N. The length of the stretched string is 20 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

उत्तर एवं व्याख्या देखें

उत्तर: Mass per unit length $\mu = 2.5 / 20 = 0.125\text{ kg/m}$. Wave speed $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = 40\text{ m/s}$. Time taken $t = \frac{L}{v} = \frac{20}{40} = 0.5\text{ seconds}$.
0.5 seconds.

Why is the sound produced by an open organ pipe richer and more pleasing than that produced by a closed organ pipe?

उत्तर एवं व्याख्या देखें

उत्तर: Because an open organ pipe produces all harmonics (both odd and even: $1f, 2f, 3f, 4f\dots$), whereas a closed organ pipe produces only odd harmonics ($1f, 3f, 5f\dots$). The presence of both even and odd harmonics creates richer musical timbre.
Open pipe contains both even and odd harmonics.

What are Beats? Two tuning forks A and B produce 5 beats per second. When fork A is filed slightly, the beat frequency increases to 7 beats/s. If frequency of B is 256 Hz, find the original frequency of A.

उत्तर एवं व्याख्या देखें

उत्तर: Beats are periodic waxing and waning of sound intensity. Initially, $f_A = 256 \pm 5 = 261\text{ Hz}$ or $251\text{ Hz}$. Filing fork A increases its frequency. If $f_A = 261$, increasing it makes beat frequency $f_A - 256$ greater than 5 (which matches 7!). If $f_A = 251$, increasing it would reduce beat frequency. Hence, original $f_A = 261\text{ Hz}$.
f_A = 261 Hz.

What is the distance between: (i) two consecutive nodes, (ii) a node and adjacent antinode in a stationary wave?

उत्तर एवं व्याख्या देखें

उत्तर: (i) Distance between two successive nodes is $\frac{\lambda}{2}$. (ii) Distance between a node and adjacent antinode is $\frac{\lambda}{4}$.
(i) λ/2, (ii) λ/4.

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