Waves NCERT Soln
Class 11 Physics : Chapter 15
Step by step NCERT Solutions for Oscillations.
Get step-by-step answers for simple harmonic motion, pendulum, and spring problems.
Question 15.1
A wave is described by the equation \( y(x, t) = 0.05 \sin (60\pi t – 2\pi x) \), where \( x \) and \( y \) are in meters and \( t \) is in seconds. What are the (a) amplitude, (b) wavelength, and (c) frequency of this wave? (d) Calculate its wave speed.
Answer & Explanation:
Compare with \( y = A \sin(\omega t – kx) \):
\( A = 0.05 \, \text{m} \)
\( k = \frac{2\pi}{\lambda} = 2\pi \Rightarrow \lambda = 1 \, \text{m} \)
\( \omega = 2\pi\nu = 60\pi \Rightarrow \nu = 30 \, \text{Hz} \)
\( v = \nu\lambda = 30 \times 1 = 30 \, \text{m/s} \) (or \( v = \frac{\omega}{k} = \frac{60\pi}{2\pi} = 30 \, \text{m/s} \))
\( A = 0.05 \, \text{m} \)
\( k = \frac{2\pi}{\lambda} = 2\pi \Rightarrow \lambda = 1 \, \text{m} \)
\( \omega = 2\pi\nu = 60\pi \Rightarrow \nu = 30 \, \text{Hz} \)
\( v = \nu\lambda = 30 \times 1 = 30 \, \text{m/s} \) (or \( v = \frac{\omega}{k} = \frac{60\pi}{2\pi} = 30 \, \text{m/s} \))
Question 15.2
For the wave \( y = 5 \sin (30t – 0.2x) \), where all quantities are in SI units, determine the phase difference between two points separated by a distance of (i) 4 m, (ii) 0.5 m.
Answer & Explanation:
Phase difference: \( \Delta\phi = k \Delta x \)
\( k = 0.2 \, \text{rad/m} \)
(i) \( \Delta x = 4 \, \text{m} \): \( \Delta\phi = 0.2 \times 4 = 0.8 \, \text{rad} \)
(ii) \( \Delta x = 0.5 \, \text{m} \): \( \Delta\phi = 0.2 \times 0.5 = 0.1 \, \text{rad} \)
\( k = 0.2 \, \text{rad/m} \)
(i) \( \Delta x = 4 \, \text{m} \): \( \Delta\phi = 0.2 \times 4 = 0.8 \, \text{rad} \)
(ii) \( \Delta x = 0.5 \, \text{m} \): \( \Delta\phi = 0.2 \times 0.5 = 0.1 \, \text{rad} \)
Question 15.3
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 \( t \) is in s. What is the direction of propagation of this wave? Also, find its amplitude, wavelength, frequency, and wave speed.
Answer & Explanation:
Form: \( y = A \sin(\omega t + kx + \phi) \) → propagates in negative x-direction.
\( A = 3.0 \, \text{cm} \)
\( \omega = 36 \, \text{rad/s} \Rightarrow \nu = \frac{\omega}{2\pi} \approx 5.73 \, \text{Hz} \)
\( k = 0.018 \, \text{rad/cm} \Rightarrow \lambda = \frac{2\pi}{k} \approx 349 \, \text{cm} \)
\( v = \frac{\omega}{k} = 2000 \, \text{cm/s} = 20 \, \text{m/s} \)
\( A = 3.0 \, \text{cm} \)
\( \omega = 36 \, \text{rad/s} \Rightarrow \nu = \frac{\omega}{2\pi} \approx 5.73 \, \text{Hz} \)
\( k = 0.018 \, \text{rad/cm} \Rightarrow \lambda = \frac{2\pi}{k} \approx 349 \, \text{cm} \)
\( v = \frac{\omega}{k} = 2000 \, \text{cm/s} = 20 \, \text{m/s} \)
Question 15.4
The equation of a plane progressive wave is given by \( y = 0.02 \sin (100\pi t – 4\pi x) \), where \( y \) and \( x \) are in meters and \( t \) is in seconds. Calculate (i) the frequency, (ii) the wavelength, (iii) the speed of the wave, and (iv) the phase difference at a given time between two points 0.25 m apart.
Answer & Explanation:
\( \omega = 100\pi \, \text{rad/s}, \quad k = 4\pi \, \text{rad/m} \)
(i) \( \nu = \frac{\omega}{2\pi} = 50 \, \text{Hz} \)
(ii) \( \lambda = \frac{2\pi}{k} = 0.5 \, \text{m} \)
(iii) \( v = \frac{\omega}{k} = 25 \, \text{m/s} \)
(iv) \( \Delta\phi = k \Delta x = 4\pi \times 0.25 = \pi \, \text{rad} = 180^\circ \)
(i) \( \nu = \frac{\omega}{2\pi} = 50 \, \text{Hz} \)
(ii) \( \lambda = \frac{2\pi}{k} = 0.5 \, \text{m} \)
(iii) \( v = \frac{\omega}{k} = 25 \, \text{m/s} \)
(iv) \( \Delta\phi = k \Delta x = 4\pi \times 0.25 = \pi \, \text{rad} = 180^\circ \)
Question 15.5
Explain the principle of superposition of waves. What is its significance?
Answer & Explanation:
Principle: When two or more waves overlap, the resultant displacement is the algebraic sum of individual displacements: \( y = y_1 + y_2 + \dots \)
Significance: Explains interference, standing waves, beats. Applies to all linear wave systems (sound, light, etc.).
Significance: Explains interference, standing waves, beats. Applies to all linear wave systems (sound, light, etc.).
Question 15.6
What are standing waves? How are they formed? Differentiate between a node and an antinode.
Answer & Explanation:
Standing waves form by superposition of two identical waves traveling in opposite directions. Energy is localized.
Node: Point of zero amplitude (permanent rest).
Antinode: Point of maximum amplitude.
Distance between consecutive nodes or antinodes = \( \lambda/2 \).
Node: Point of zero amplitude (permanent rest).
Antinode: Point of maximum amplitude.
Distance between consecutive nodes or antinodes = \( \lambda/2 \).
Question 15.7
A wire of length 1.5 m and mass 6 g is under a tension of 80 N. What is the speed of a transverse wave on the wire? If the wire is fixed at both ends, what is its fundamental frequency of vibration?
Answer & Explanation:
\( \mu = \frac{m}{L} = \frac{0.006}{1.5} = 0.004 \, \text{kg/m} \)
\( v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{80}{0.004}} = 100\sqrt{2} \approx 141.4 \, \text{m/s} \)
\( f_1 = \frac{v}{2L} = \frac{141.4}{3} \approx 47.1 \, \text{Hz} \)
\( v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{80}{0.004}} = 100\sqrt{2} \approx 141.4 \, \text{m/s} \)
\( f_1 = \frac{v}{2L} = \frac{141.4}{3} \approx 47.1 \, \text{Hz} \)
Question 15.8
A pipe 30 cm long is closed at one end. The velocity of sound in air is 330 m/s. Find the fundamental frequency of vibration of the air column in the pipe. Which harmonic will have a frequency of 2200 Hz?
Answer & Explanation:
Closed pipe: \( f_n = \frac{nv}{4L}, \quad n = 1, 3, 5, \dots \)
\( f_1 = \frac{330}{4 \times 0.3} = 275 \, \text{Hz} \)
For \( f = 2200 \, \text{Hz} \): \( n = \frac{2200}{275} = 8 \) (not odd) → not a harmonic.
\( f_1 = \frac{330}{4 \times 0.3} = 275 \, \text{Hz} \)
For \( f = 2200 \, \text{Hz} \): \( n = \frac{2200}{275} = 8 \) (not odd) → not a harmonic.
Question 15.9
Two sitar strings A and B are slightly out of tune and produce beats of frequency 5 Hz. The tension in string A is increased slightly. It is observed that the beat frequency decreases. If the original frequency of A is 425 Hz, what was the original frequency of B?
Answer & Explanation:
Beat frequency = \( |f_A - f_B| = 5 \, \text{Hz} \).
\( f_A = 425 \, \text{Hz} \) → \( f_B = 420 \, \text{Hz} \) or \( 430 \, \text{Hz} \).
Increasing \( f_A \) makes it closer to \( 430 \, \text{Hz} \), decreasing beat frequency.
So \( f_B = \mathbf{430 \, \text{Hz}} \).
\( f_A = 425 \, \text{Hz} \) → \( f_B = 420 \, \text{Hz} \) or \( 430 \, \text{Hz} \).
Increasing \( f_A \) makes it closer to \( 430 \, \text{Hz} \), decreasing beat frequency.
So \( f_B = \mathbf{430 \, \text{Hz}} \).
Question 15.10
Explain why (or how):
a) A sound wave is classified as a longitudinal wave.
b) Bats can ascertain distances, directions, nature, and sizes of obstacles without seeing them.
c) A violin note and a sitar note may have the same frequency, yet we can distinguish between the two.
d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases.
a) A sound wave is classified as a longitudinal wave.
b) Bats can ascertain distances, directions, nature, and sizes of obstacles without seeing them.
c) A violin note and a sitar note may have the same frequency, yet we can distinguish between the two.
d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases.
Answer & Explanation:
a) Particle oscillation parallel to propagation → compressions & rarefactions.
b) Echolocation using ultrasonic waves and analyzing reflections.
c) Different timbre due to different harmonic content.
d) Solids have shear modulus (transverse) and bulk modulus (longitudinal); gases lack shear modulus.
b) Echolocation using ultrasonic waves and analyzing reflections.
c) Different timbre due to different harmonic content.
d) Solids have shear modulus (transverse) and bulk modulus (longitudinal); gases lack shear modulus.
Question 15.11
A train, standing at the outer signal of a railway station, blows a whistle of frequency 500 Hz in still air. The speed of sound in still air is 340 m/s.
(i) What is the frequency of the whistle for a platform observer when the train approaches the platform with a speed of 20 m/s?
(ii) What is the frequency heard by the same observer when the train recedes from the platform with the same speed?
(i) What is the frequency of the whistle for a platform observer when the train approaches the platform with a speed of 20 m/s?
(ii) What is the frequency heard by the same observer when the train recedes from the platform with the same speed?
Answer & Explanation:
Doppler effect (source moving, observer stationary):
\( f' = f \frac{v}{v \mp v_s} \)
(i) Approaching: \( f' = 500 \times \frac{340}{340-20} = 531.25 \, \text{Hz} \)
(ii) Receding: \( f'' = 500 \times \frac{340}{340+20} \approx 472.2 \, \text{Hz} \)
\( f' = f \frac{v}{v \mp v_s} \)
(i) Approaching: \( f' = 500 \times \frac{340}{340-20} = 531.25 \, \text{Hz} \)
(ii) Receding: \( f'' = 500 \times \frac{340}{340+20} \approx 472.2 \, \text{Hz} \)
Question 15.12
The displacement of a particle in a wave is given by \( y = 0.2 \sin (0.4\pi (50t – x)) \). Here \( x \) and \( y \) are in cm and \( t \) in s. Calculate the wave velocity and the maximum particle velocity. Are they equal? If not, why?
Answer & Explanation:
\( y = 0.2 \sin(20\pi t - 0.4\pi x) \)
\( v_{\text{wave}} = \frac{\omega}{k} = \frac{20\pi}{0.4\pi} = 50 \, \text{cm/s} \)
\( v_{\text{particle,max}} = A\omega = 0.2 \times 20\pi \approx 12.57 \, \text{cm/s} \)
Not equal: wave velocity is phase speed; particle velocity is oscillation speed.
\( v_{\text{wave}} = \frac{\omega}{k} = \frac{20\pi}{0.4\pi} = 50 \, \text{cm/s} \)
\( v_{\text{particle,max}} = A\omega = 0.2 \times 20\pi \approx 12.57 \, \text{cm/s} \)
Not equal: wave velocity is phase speed; particle velocity is oscillation speed.
Question 15.13
A steel rod 1 m long is clamped at its midpoint. It is set into longitudinal vibrations. Given the density of steel = 8000 kg/m³ and Young's modulus = 2 × 10¹¹ N/m², what is its fundamental frequency?
Answer & Explanation:
Clamped at midpoint → \( L = \lambda/2 \Rightarrow \lambda = 2L = 2 \, \text{m} \)
\( v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{2 \times 10^{11}}{8000}} = 5000 \, \text{m/s} \)
\( f = \frac{v}{\lambda} = \frac{5000}{2} = 2500 \, \text{Hz} \)
\( v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{2 \times 10^{11}}{8000}} = 5000 \, \text{m/s} \)
\( f = \frac{v}{\lambda} = \frac{5000}{2} = 2500 \, \text{Hz} \)
Question 15.14
Two waves of equal frequency and amplitude traveling in opposite directions on a string superimpose to produce standing waves. If the amplitude of each wave is 2 cm, what is the amplitude at an antinode? What is the amplitude at a node?
Answer & Explanation:
Resultant amplitude = \( 2A |\cos(kx)| \)
Antinode: \( \cos(kx) = \pm 1 \) → Amplitude = \( 2 \times 2 = 4 \, \text{cm} \)
Node: \( \cos(kx) = 0 \) → Amplitude = \( 0 \, \text{cm} \)
Antinode: \( \cos(kx) = \pm 1 \) → Amplitude = \( 2 \times 2 = 4 \, \text{cm} \)
Node: \( \cos(kx) = 0 \) → Amplitude = \( 0 \, \text{cm} \)
Question 15.15
A tuning fork of unknown frequency produces 4 beats per second with another tuning fork of frequency 256 Hz. When the prongs of the first fork are loaded with a little wax, the beat frequency is found to decrease. What is the original frequency of the first fork?
Answer & Explanation:
Let unknown frequency = \( f \). \( |f - 256| = 4 \) → \( f = 260 \) or \( 252 \, \text{Hz} \).
Wax decreases frequency. If \( f = 260 \), decrease brings it closer to 256 → beat frequency decreases.
Therefore \( f = \mathbf{260 \, \text{Hz}} \).
Wax decreases frequency. If \( f = 260 \), decrease brings it closer to 256 → beat frequency decreases.
Therefore \( f = \mathbf{260 \, \text{Hz}} \).
Question 15.16
What is the difference between the speed of a transverse wave on a string and the speed of a sound wave in air? Derive the expression for the speed of a transverse wave on a stretched string.
Answer & Explanation:
Difference:
• String: \( v = \sqrt{T/\mu} \) (depends on tension and linear density)
• Sound in air: \( v = \sqrt{\gamma P/\rho} \) (depends on gas properties)
Derivation: From wave equation \( \frac{\partial^2 y}{\partial t^2} = \frac{T}{\mu} \frac{\partial^2 y}{\partial x^2} \) → \( v^2 = \frac{T}{\mu} \) → \( v = \sqrt{\frac{T}{\mu}} \)
• String: \( v = \sqrt{T/\mu} \) (depends on tension and linear density)
• Sound in air: \( v = \sqrt{\gamma P/\rho} \) (depends on gas properties)
Derivation: From wave equation \( \frac{\partial^2 y}{\partial t^2} = \frac{T}{\mu} \frac{\partial^2 y}{\partial x^2} \) → \( v^2 = \frac{T}{\mu} \) → \( v = \sqrt{\frac{T}{\mu}} \)
Question 15.17
A stone dropped from the top of a tower 300 m high splashes into a pond at the base of the tower. When is the splash heard at the top? (Given g = 10 m/s², speed of sound in air = 340 m/s)
Answer & Explanation:
Time to fall: \( t_1 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{600}{10}} = \sqrt{60} \approx 7.746 \, \text{s} \)
Sound travel time: \( t_2 = \frac{h}{v} = \frac{300}{340} \approx 0.882 \, \text{s} \)
Total: \( T = t_1 + t_2 \approx 8.628 \, \text{s} \)
Sound travel time: \( t_2 = \frac{h}{v} = \frac{300}{340} \approx 0.882 \, \text{s} \)
Total: \( T = t_1 + t_2 \approx 8.628 \, \text{s} \)
Question 15.18
A wave traveling along a string is described by \( y = 0.005 \sin (80x – 3t) \) in SI units. Calculate (a) the amplitude, (b) the wavelength, (c) the period, and (d) the frequency of the wave.
Answer & Explanation:
\( A = 0.005 \, \text{m} \)
\( k = 80 \, \text{rad/m} \Rightarrow \lambda = \frac{2\pi}{k} \approx 0.0785 \, \text{m} \)
\( \omega = 3 \, \text{rad/s} \Rightarrow T = \frac{2\pi}{\omega} \approx 2.094 \, \text{s} \)
\( f = \frac{1}{T} \approx 0.477 \, \text{Hz} \)
\( k = 80 \, \text{rad/m} \Rightarrow \lambda = \frac{2\pi}{k} \approx 0.0785 \, \text{m} \)
\( \omega = 3 \, \text{rad/s} \Rightarrow T = \frac{2\pi}{\omega} \approx 2.094 \, \text{s} \)
\( f = \frac{1}{T} \approx 0.477 \, \text{Hz} \)
Question 15.19
For a traveling harmonic wave \( y = 2.0 \cos 2\pi(10t – 0.008x + 0.35) \), where \( x \) and \( y \) are in cm and \( t \) is in s. What is the phase difference between oscillatory motion at two points separated by a distance of (a) 4 m, (b) 0.5 m, (c) \( \lambda/2 \), (d) \( 3\lambda/4 \)?
Answer & Explanation:
\( k = 0.016\pi \, \text{rad/cm} \)
(a) \( \Delta x = 400 \, \text{cm} \): \( \Delta\phi = k \Delta x = 6.4\pi \equiv 0.4\pi \, \text{rad} \)
(b) \( \Delta x = 50 \, \text{cm} \): \( \Delta\phi = 0.8\pi \, \text{rad} \)
(c) \( \Delta\phi = \pi \, \text{rad} \)
(d) \( \Delta\phi = 3\pi/2 \, \text{rad} \)
(a) \( \Delta x = 400 \, \text{cm} \): \( \Delta\phi = k \Delta x = 6.4\pi \equiv 0.4\pi \, \text{rad} \)
(b) \( \Delta x = 50 \, \text{cm} \): \( \Delta\phi = 0.8\pi \, \text{rad} \)
(c) \( \Delta\phi = \pi \, \text{rad} \)
(d) \( \Delta\phi = 3\pi/2 \, \text{rad} \)
Question 15.20
The transverse displacement of a string is given by \( y(x, t) = 0.06 \sin (\frac{2\pi x}{3}) \cos (120\pi t) \). All quantities are in SI units. (a) Is this a traveling or standing wave? (b) What are its amplitude, wavelength, frequency, and wave speed? (c) Identify nodes and antinodes.
Answer & Explanation:
(a) Standing wave
(b) Max amplitude = 0.06 m, \( k = \frac{2\pi}{3} \Rightarrow \lambda = 3 \, \text{m} \), \( \omega = 120\pi \Rightarrow f = 60 \, \text{Hz} \), \( v = f\lambda = 180 \, \text{m/s} \)
(c) Nodes: \( x = \frac{3n}{2} \) m (n=0,1,2...); Antinodes: \( x = \frac{3(2n+1)}{4} \) m
(b) Max amplitude = 0.06 m, \( k = \frac{2\pi}{3} \Rightarrow \lambda = 3 \, \text{m} \), \( \omega = 120\pi \Rightarrow f = 60 \, \text{Hz} \), \( v = f\lambda = 180 \, \text{m/s} \)
(c) Nodes: \( x = \frac{3n}{2} \) m (n=0,1,2...); Antinodes: \( x = \frac{3(2n+1)}{4} \) m
Question 15.21
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 50 Hz. The mass of the wire is 3 × 10⁻² kg and its linear density is 4 × 10⁻² kg/m. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?
Answer & Explanation:
\( L = \frac{m}{\mu} = \frac{0.03}{0.04} = 0.75 \, \text{m} \)
(a) \( v = 2L f_1 = 2 \times 0.75 \times 50 = 75 \, \text{m/s} \)
(b) \( T = \mu v^2 = 0.04 \times 75^2 = 225 \, \text{N} \)
(a) \( v = 2L f_1 = 2 \times 0.75 \times 50 = 75 \, \text{m/s} \)
(b) \( T = \mu v^2 = 0.04 \times 75^2 = 225 \, \text{N} \)
Question 15.22
A man standing symmetrically between two parallel cliffs fires a gun. He hears the first echo after 2 s and the second echo 1 s after the first. Calculate the distance between the cliffs. (Speed of sound in air = 340 m/s)
Answer & Explanation:
Let distances to cliffs be \( x \) and \( y \) (\( x < y \)):
\( 2x = v t_1 = 340 \times 2 \Rightarrow x = 340 \, \text{m} \)
\( 2y = v (t_1 + t_2) = 340 \times 3 \Rightarrow y = 510 \, \text{m} \)
Distance between cliffs = \( x + y = 850 \, \text{m} \)
\( 2x = v t_1 = 340 \times 2 \Rightarrow x = 340 \, \text{m} \)
\( 2y = v (t_1 + t_2) = 340 \times 3 \Rightarrow y = 510 \, \text{m} \)
Distance between cliffs = \( x + y = 850 \, \text{m} \)
Question 15.23
A bat emits ultrasonic sound of frequency 120 kHz in air. If this 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 = 1480 m/s)
Answer & Explanation:
\( f = 1.2 \times 10^5 \, \text{Hz} \)
(a) Reflected (air): \( \lambda_{\text{air}} = \frac{340}{1.2 \times 10^5} \approx 2.83 \, \text{mm} \)
(b) Transmitted (water): \( \lambda_{\text{water}} = \frac{1480}{1.2 \times 10^5} \approx 12.33 \, \text{mm} \)
(a) Reflected (air): \( \lambda_{\text{air}} = \frac{340}{1.2 \times 10^5} \approx 2.83 \, \text{mm} \)
(b) Transmitted (water): \( \lambda_{\text{water}} = \frac{1480}{1.2 \times 10^5} \approx 12.33 \, \text{mm} \)
Question 15.24
State the laws of transverse vibrations of a stretched string (Mersenne's laws).
Answer & Explanation:
1. Law of length: \( f \propto 1/L \)
2. Law of tension: \( f \propto \sqrt{T} \)
3. Law of density: \( f \propto 1/\sqrt{\mu} \)
(All for constant other parameters)
2. Law of tension: \( f \propto \sqrt{T} \)
3. Law of density: \( f \propto 1/\sqrt{\mu} \)
(All for constant other parameters)
Question 15.25
Explain the formation of beats analytically. Show that the beat frequency is equal to the absolute difference in the frequencies of the two superposing waves.
Answer & Explanation:
Let \( y_1 = A \sin\omega_1 t, \quad y_2 = A \sin\omega_2 t \).
\( y = y_1 + y_2 = 2A \cos\left(\frac{\omega_1 - \omega_2}{2}t\right) \sin\left(\frac{\omega_1 + \omega_2}{2}t\right) \)
Amplitude varies as \( 2A \cos(\omega_{\text{mod}} t) \) where \( \omega_{\text{mod}} = \frac{|\omega_1 - \omega_2|}{2} \).
Beat period \( T_b = \frac{\pi}{\omega_{\text{mod}}} \) → Beat frequency \( f_b = \frac{1}{T_b} = \frac{|\omega_1 - \omega_2|}{2\pi} = |f_1 - f_2| \).
\( y = y_1 + y_2 = 2A \cos\left(\frac{\omega_1 - \omega_2}{2}t\right) \sin\left(\frac{\omega_1 + \omega_2}{2}t\right) \)
Amplitude varies as \( 2A \cos(\omega_{\text{mod}} t) \) where \( \omega_{\text{mod}} = \frac{|\omega_1 - \omega_2|}{2} \).
Beat period \( T_b = \frac{\pi}{\omega_{\text{mod}}} \) → Beat frequency \( f_b = \frac{1}{T_b} = \frac{|\omega_1 - \omega_2|}{2\pi} = |f_1 - f_2| \).
Question 15.26
What is the Doppler effect? Derive an expression for the apparent frequency when a source of sound moves towards a stationary observer.
Answer & Explanation:
Doppler effect: Change in observed frequency due to relative motion.
Derivation (source moving toward observer):
Apparent wavelength \( \lambda' = \frac{v - v_s}{f} \)
\( f' = \frac{v}{\lambda'} = f \frac{v}{v - v_s} \)
Derivation (source moving toward observer):
Apparent wavelength \( \lambda' = \frac{v - v_s}{f} \)
\( f' = \frac{v}{\lambda'} = f \frac{v}{v - v_s} \)
Question 15.27
Discuss briefly the analytical treatment of stationary waves in a stretched string fixed at both ends. Show that the frequencies of normal modes are integral multiples of the fundamental frequency.
Answer & Explanation:
Boundary conditions: \( y(0,t) = y(L,t) = 0 \)
Solution: \( y_n = A_n \sin(k_n x) \cos(\omega_n t) \) with \( k_n L = n\pi \)
\( \lambda_n = \frac{2L}{n}, \quad f_n = \frac{v}{\lambda_n} = \frac{nv}{2L} = n f_1 \) where \( f_1 = \frac{v}{2L} \)
Solution: \( y_n = A_n \sin(k_n x) \cos(\omega_n t) \) with \( k_n L = n\pi \)
\( \lambda_n = \frac{2L}{n}, \quad f_n = \frac{v}{\lambda_n} = \frac{nv}{2L} = n f_1 \) where \( f_1 = \frac{v}{2L} \)
📘 Exam Preparation Tip:
This comprehensive exercise set covers all key concepts from NCERT Class 11 Physics Chapter 15: Waves. Students will master wave classification, mathematical representation of progressive waves, superposition principle, standing waves formation, beat phenomena, and Doppler effect applications. Through detailed solutions, they'll learn to calculate wave speed in different media, determine resonant frequencies, solve interference problems, and apply wave concepts to real-world scenarios like earthquake detection and ultrasonic navigation. These exercises strengthen problem-solving skills for board exams and competitive tests.