CLASS 11 PHYSICS CHAPTER 14 – WAVES
NCERT CLASS 11 PHYSICS
CHAPTER 14 – WAVES
IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS
1. What is a wave?
Answer:
A wave is a disturbance that travels through a medium or space and transfers energy from one place to another without the bulk transfer of matter.
In a mechanical wave, particles of the medium oscillate about their equilibrium positions.
Examples:
Waves travelling along a stretched string.
Sound waves travelling through air.
Water surface waves.
2. What is a mechanical wave?
Answer:
A mechanical wave is a disturbance that requires a material medium for its propagation.
It travels through the interaction of particles of the medium.
Examples:
Sound waves.
Waves on a stretched string.
Water waves.
Mechanical waves cannot travel through a vacuum.
3. What are transverse waves?
Answer:
Transverse waves are waves in which particles of the medium vibrate perpendicular to the direction of propagation of the wave.
Examples:
Waves on a stretched string.
Electromagnetic waves.
In a transverse wave, crests and troughs are formed.
4. What are longitudinal waves?
Answer:
Longitudinal waves are waves in which particles of the medium vibrate parallel to the direction of propagation of the wave.
Examples:
Sound waves in air.
Compression waves in a spring.
Longitudinal waves consist of compressions and rarefactions.
5. Differentiate between transverse and longitudinal waves.
Answer:
Transverse Waves:
Particles vibrate perpendicular to the direction of wave propagation.
They have crests and troughs.
Mechanical transverse waves require a medium capable of providing a shear restoring force.
Example: Waves on a stretched string.
Longitudinal Waves:
Particles vibrate parallel to the direction of wave propagation.
They have compressions and rarefactions.
They can travel through solids, liquids and gases.
Example: Sound waves in air.
6. What is amplitude of a wave?
Answer:
Amplitude is the maximum displacement of a particle of the medium from its mean or equilibrium position.
It is represented by A or a.
SI unit: metre (m).
7. What is wavelength?
Answer:
Wavelength is the distance between two consecutive particles of a wave that are in the same phase.
For example:
Distance between two consecutive crests.
Distance between two consecutive troughs.
Distance between two consecutive compressions.
It is represented by λ (lambda).
SI unit: metre (m).
8. Define frequency and time period of a wave.
Answer:
Frequency:
The number of complete oscillations made by a particle in one second is called frequency.
It is represented by ν.
SI unit: hertz (Hz).
Time Period:
The time taken by a particle to complete one oscillation is called the time period.
It is represented by T.
SI unit: second (s).
Relation:
ν = 1/T
9. What is the speed of a wave?
Answer:
Wave speed is the distance travelled by a wave disturbance per unit time.
It is represented by v.
The relation between wave speed, frequency and wavelength is:
v = νλ
where:
v = Wave speed
ν = Frequency
λ = Wavelength
SI unit: metre per second (m/s).
10. What is angular frequency?
Answer:
Angular frequency is the rate of change of phase of an oscillating particle with time.
It is represented by ω.
ω = 2πν
or
ω = 2π/T
SI unit: radian per second (rad/s).
11. What is wave number?
Answer:
Wave number is the spatial rate of change of phase of a wave.
It is represented by k.
k = 2π/λ
where λ is the wavelength.
SI unit: radian per metre (rad/m).
12. What is a progressive wave?
Answer:
A progressive wave is a wave that travels continuously through a medium and transfers energy from one point to another.
In a progressive wave:
All particles oscillate with the same frequency.
All particles have the same amplitude in an ideal uniform medium.
The phase of oscillation changes continuously from one point to another.
Energy is transferred along the direction of propagation.
13. What is the principle of superposition of waves?
Answer:
The principle of superposition states that when two or more waves travel through a medium simultaneously, the resultant displacement of any particle is equal to the algebraic sum of the individual displacements produced by each wave.
Mathematically:
y = y₁ + y₂ + y₃ + ...
14. What is interference of waves?
Answer:
Interference is the phenomenon in which two or more waves overlap and combine to produce a resultant wave.
It may produce:
Constructive interference.
Destructive interference.
15. What is constructive interference?
Answer:
Constructive interference occurs when two waves meet in the same phase and their displacements add together.
The resultant amplitude becomes maximum.
For two waves of equal amplitude A:
Resultant amplitude = 2A
Condition:
Path difference = nλ
Phase difference = 2nπ
where n = 0, 1, 2, 3, ...
16. What is destructive interference?
Answer:
Destructive interference occurs when two waves meet in opposite phases and their displacements cancel each other.
For two waves of equal amplitude, the resultant amplitude becomes zero.
Condition:
Path difference = (2n + 1)λ/2
Phase difference = (2n + 1)π
17. What are stationary waves?
Answer:
Stationary waves are formed when two waves of equal frequency and amplitude travel in opposite directions and superpose.
They produce fixed points called nodes and antinodes.
In a stationary wave:
Nodes and antinodes remain at fixed positions.
There is no net transfer of energy along the medium.
The amplitude varies from point to point.
18. What are nodes and antinodes?
Answer:
Node:
A node is a point in a stationary wave where the displacement of the particle is always zero.
Antinode:
An antinode is a point where the displacement amplitude is maximum.
Distance between two consecutive nodes = λ/2.
Distance between two consecutive antinodes = λ/2.
Distance between a node and its nearest antinode = λ/4.
19. What are beats?
Answer:
Beats are the periodic variations in loudness heard when two sound waves of slightly different frequencies travel together.
The number of beats heard per second is called beat frequency.
Beat frequency:
νbeat = |ν₁ − ν₂|
20. What is resonance in sound?
Answer:
Resonance occurs when the frequency of an external periodic force becomes equal or nearly equal to the natural frequency of a system.
In an air column, resonance produces a large amplitude of vibration when the frequency of the sound source matches one of the natural frequencies of the air column.
21. What is the Doppler effect?
Answer:
The Doppler effect is the apparent change in frequency of a wave due to relative motion between the source and the observer.
For example, the pitch of a moving ambulance siren appears higher when it approaches an observer and lower when it moves away.
22. Why can sound not travel through a vacuum?
Answer:
Sound is a mechanical wave. It requires a material medium for propagation.
In a vacuum, there are no particles to transmit the disturbance.
Therefore, sound cannot travel through a vacuum.
EXERCISE QUESTIONS AND ANSWERS
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 disturbance is produced at one end of the string, how long does it take to reach the other end?
Answer:
Given:
Mass of string, m = 2.50 kg
Length of string, L = 20.0 m
Tension, T = 200 N
Step 1: Calculate mass per unit length.
Linear mass density:
μ = m/L
μ = 2.50/20
μ = 0.125 kg/m
Step 2: Calculate the speed of the transverse wave.
Formula:
v = √(T/μ)
Substituting:
v = √(200/0.125)
v = √1600
v = 40 m/s
Step 3: Calculate the time taken.
Time = Distance/Speed
t = L/v
t = 20/40
t = 0.5 s
Final Answer:
The disturbance takes 0.5 seconds to reach the other end.
Q14.2. A stone is dropped from the top of a tower of height 300 m. It splashes into the water of a pond near the base of the tower. When will the splash be heard at the top? The speed of sound in air is 340 m/s. Take g = 9.8 m/s².
Answer:
Given:
Height, h = 300 m
Speed of sound, v = 340 m/s
Acceleration due to gravity, g = 9.8 m/s²
Step 1: Calculate the time taken by the stone to reach the water.
Using:
h = (1/2)gt₁²
Therefore,
t₁ = √(2h/g)
t₁ = √(2 × 300/9.8)
t₁ = √61.22
t₁ = 7.82 s
Step 2: Calculate the time taken by sound to reach the top.
t₂ = Distance/Speed
t₂ = 300/340
t₂ = 0.88 s
Step 3: Calculate the total time.
Total time = t₁ + t₂
= 7.82 + 0.88
= 8.70 s
Final Answer:
The splash will be heard after approximately 8.7 seconds.
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 °C, which is 343 m/s?
Answer:
Given:
Length of wire, L = 12.0 m
Mass of wire, m = 2.10 kg
Wave speed, v = 343 m/s
Step 1: Calculate linear mass density.
μ = m/L
μ = 2.10/12.0
μ = 0.175 kg/m
Step 2: Use the wave speed formula.
v = √(T/μ)
Squaring both sides:
v² = T/μ
Therefore,
T = μv²
Substituting:
T = 0.175 × 343²
T = 0.175 × 117649
T = 20588.6 N
Final Answer:
Required tension = approximately 2.06 × 10⁴ N.
Q14.4. Use the formula v = √(γP/ρ) to explain why the speed of sound in air is (a) independent of pressure, (b) increases with temperature, and (c) increases with humidity.
Answer:
The speed of sound in a gas is:
v = √(γP/ρ)
where:
γ = Ratio of specific heats.
P = Pressure.
ρ = Density.
Using the ideal gas equation:
P = ρRT/M
Therefore,
P/ρ = RT/M
Substituting:
v = √(γRT/M)
(a) Why is the speed of sound independent of pressure?
From:
v = √(γRT/M)
The expression does not contain pressure.
At constant temperature, pressure and density change in the same proportion.
Therefore, the ratio P/ρ remains constant.
Hence, the speed of sound is independent of pressure at a fixed temperature.
(b) Why does the speed of sound increase with temperature?
From:
v = √(γRT/M)
At constant composition:
v ∝ √T
Therefore, the speed of sound increases when the absolute temperature increases.
(c) Why does the speed of sound increase with humidity?
Water vapour has a lower molar mass than the main constituents of dry air.
As humidity increases, the average molar mass of air decreases.
From:
v = √(γRT/M)
A decrease in M increases the speed of sound.
Final Answer:
(a) At constant temperature, speed is independent of pressure.
(b) Speed increases with temperature.
(c) Speed increases with humidity.
Q14.5. A travelling wave in one dimension is represented by a function y = f(x, t), where x and t appear in the combination x − vt or x + vt. Is the converse true? Examine whether the following functions can represent a travelling wave:
(a) (x − vt)²
(b) log[(x + vt)/x₀]
(c) 1/(x + vt)
Answer:
A travelling wave moving along the positive or negative x-direction can be represented by:
y = f(x − vt)
or
y = f(x + vt)
The converse requires care: the function must also be physically meaningful over the region in which the wave is defined.
(a) y = (x − vt)²
This function depends only on the combination (x − vt).
It is finite for every finite value of x and t.
Therefore, it represents a mathematical travelling-wave profile moving in the positive x-direction.
It is not a localised pulse because its value grows without bound as |x − vt| increases.
(b) y = log[(x + vt)/x₀]
The logarithm is defined only when:
(x + vt)/x₀ > 0
It becomes undefined at x + vt = 0 and is not defined over the entire real x-axis.
Therefore, it cannot represent a physically valid travelling wave over the entire line without restricting its domain.
(c) y = 1/(x + vt)
This function becomes infinite at:
x + vt = 0
Therefore, it is singular at that point and cannot represent a physically valid travelling wave over the entire real x-axis.
Final Answer:
(a) It is a mathematical travelling-wave profile, although it is not a localised finite-energy pulse.
(b) It is not valid over the entire real x-axis.
(c) It is not valid over the entire real x-axis because of its singularity.
Q14.6. A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, calculate the wavelength of (a) the reflected sound and (b) the transmitted sound. The speed of sound in air is 340 m/s and in water is 1486 m/s.
Answer:
Given:
Frequency, ν = 1000 kHz
ν = 1000 × 1000
ν = 10⁶ Hz
Speed in air = 340 m/s
Speed in water = 1486 m/s
The frequency of a wave remains unchanged when it is reflected or transmitted across a boundary.
(a) Wavelength of reflected sound
The reflected sound travels in air.
Formula:
λ = v/ν
λ = 340/10⁶
λ = 3.4 × 10⁻⁴ m
(b) Wavelength of transmitted sound
The transmitted sound travels in water.
λ = v/ν
λ = 1486/10⁶
λ = 1.486 × 10⁻³ m
Final Answer:
(a) Wavelength of reflected sound = 3.4 × 10⁻⁴ m.
(b) Wavelength of transmitted sound = 1.486 × 10⁻³ m.
Q14.7. A hospital uses an ultrasonic scanner to locate tumours in tissue. What is the wavelength of sound in the tissue if the speed of sound is 1.7 km/s and the operating frequency of the scanner is 4.2 MHz?
Answer:
Given:
Speed, v = 1.7 km/s
v = 1.7 × 10³ m/s
Frequency, ν = 4.2 MHz
ν = 4.2 × 10⁶ Hz
Formula:
λ = v/ν
Substituting:
λ = (1.7 × 10³)/(4.2 × 10⁶)
λ = 0.4048 × 10⁻³ m
λ = 4.05 × 10⁻⁴ m
Final Answer:
Wavelength = 4.05 × 10⁻⁴ m or approximately 0.405 mm.
Q14.8. A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin(36t + 0.018x + π/4), where x and y are in centimetres and t is in seconds. The positive direction of x is from left to right. (a) Is this a travelling wave or a stationary wave? If it is travelling, what are its speed and direction of propagation? (b) What are its amplitude and frequency? (c) What is its initial phase at the origin? (d) What is the least distance between two successive crests?
Answer:
Given wave equation:
y(x, t) = 3.0 sin(36t + 0.018x + π/4)
Compare it with:
y = A sin(ωt + kx + φ)
Therefore:
Amplitude, A = 3.0 cm
Angular frequency, ω = 36 rad/s
Wave number, k = 0.018 rad/cm
Initial phase, φ = π/4
(a) Nature, speed and direction of propagation
The wave equation contains the combination:
ωt + kx
This represents a travelling wave moving in the negative x-direction.
Wave speed:
v = ω/k
v = 36/0.018
v = 2000 cm/s
Converting into metres per second:
v = 2000/100
v = 20 m/s
(b) Amplitude and frequency
Amplitude:
A = 3.0 cm
Frequency:
ν = ω/2π
ν = 36/(2 × 3.14)
ν ≈ 5.73 Hz
(c) Initial phase at the origin
At x = 0 and t = 0:
Initial phase = π/4 rad
(d) Distance between successive crests
The distance between two consecutive crests is one wavelength.
λ = 2π/k
λ = 2 × 3.14/0.018
λ ≈ 349 cm
λ ≈ 3.49 m
Final Answer:
(a) Travelling wave moving in the negative x-direction; speed = 20 m/s.
(b) Amplitude = 3.0 cm; frequency = 5.73 Hz.
(c) Initial phase = π/4 rad.
(d) Wavelength = 3.49 m approximately.
Q14.9. For the wave described in Exercise 14.8, plot the displacement y versus time t graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in a travelling wave differ from one point to another: amplitude, frequency or phase?
Answer:
The given wave equation is:
y(x, t) = 3.0 sin(36t + 0.018x + π/4)
At x = 0 cm
y = 3.0 sin(36t + π/4)
At x = 2 cm
y = 3.0 sin(36t + 0.036 + π/4)
At x = 4 cm
y = 3.0 sin(36t + 0.072 + π/4)
Each equation represents a sine curve.
Therefore, all three displacement-time graphs are sinusoidal curves.
The amplitude of all three graphs is:
A = 3.0 cm
Their angular frequency is:
ω = 36 rad/s
Their frequency is:
ν = 36/2π ≈ 5.73 Hz
Their time period is:
T = 2π/36 ≈ 0.175 s
The phase changes with position because the term 0.018x changes.
Final Answer:
All three graphs are sinusoidal curves.
All points have the same amplitude of 3.0 cm.
All points have the same frequency of approximately 5.73 Hz.
The phase is different at different positions.
Thus, in a travelling wave, particles differ in phase, not in amplitude or frequency, for the given uniform wave.
Q14.10. For the travelling harmonic wave y(x, t) = 2.0 cos 2π(10t − 0.0080x + 0.35), where x and y are in centimetres and t is in seconds, calculate the phase difference between the oscillatory motion of two points separated by a distance of (a) 4 m, (b) 0.5 m, (c) λ/2 and (d) 3λ/4.
Answer:
Given:
y(x, t) = 2.0 cos 2π(10t − 0.0080x + 0.35)
Expanding:
y = 2.0 cos(20πt − 0.016πx + 0.70π)
Wave number:
k = 0.016π rad/cm
Phase difference:
Δφ = kΔx
(a) Separation = 4 m
Converting:
4 m = 400 cm
Δφ = 0.016π × 400
Δφ = 6.4π rad
Equivalent phase difference within one complete cycle:
6.4π − 6π = 0.4π rad
(b) Separation = 0.5 m
0.5 m = 50 cm
Δφ = 0.016π × 50
Δφ = 0.8π rad
(c) Separation = λ/2
Phase difference for one wavelength = 2π.
Therefore,
Δφ = 2π × 1/2
Δφ = π rad
(d) Separation = 3λ/4
Δφ = 2π × 3/4
Δφ = 3π/2 rad
Final Answer:
(a) 6.4π rad, equivalent to 0.4π rad.
(b) 0.8π rad.
(c) π rad.
(d) 3π/2 rad.
Q14.11. The transverse displacement of a string clamped at both ends is given by y(x, t) = 0.06 sin(2πx/3) cos(120πt), where x and y are in metres and t is in seconds. The length of the string is 1.5 m and its mass is 3.0 × 10⁻² kg. Answer the following: (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 are the wavelength, frequency and speed of each wave? (c) Determine the tension in the string.
Answer:
Given:
y(x, t) = 0.06 sin(2πx/3) cos(120πt)
Length, L = 1.5 m
Mass, m = 3.0 × 10⁻² kg
(a) Travelling wave or stationary wave?
The equation can be written as:
y = 0.06 sin(kx) cos(ωt)
The position-dependent and time-dependent terms are multiplied separately.
Therefore, it represents a stationary wave.
(b) Wavelength, frequency and speed
Comparing:
k = 2π/3 rad/m
ω = 120π rad/s
Wavelength:
λ = 2π/k
λ = 2π/(2π/3)
λ = 3 m
Frequency:
ν = ω/2π
ν = 120π/2π
ν = 60 Hz
Wave speed:
v = νλ
v = 60 × 3
v = 180 m/s
The stationary wave is formed by two waves travelling in opposite directions, each having the same wavelength, frequency and speed.
(c) Calculate the tension
Linear mass density:
μ = m/L
μ = (3.0 × 10⁻²)/1.5
μ = 0.02 kg/m
Speed of transverse wave:
v = √(T/μ)
Squaring:
T = μv²
T = 0.02 × 180²
T = 0.02 × 32400
T = 648 N
Final Answer:
(a) It is a stationary wave.
(b) Wavelength = 3 m, frequency = 60 Hz, speed = 180 m/s.
(c) Tension = 648 N.
Q14.12. (i) For the wave on a string described in Exercise 14.11, do all the points on the string oscillate with the same (a) frequency, (b) phase and (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?
Answer:
The given wave equation is:
y(x, t) = 0.06 sin(2πx/3) cos(120πt)
(i) Do all points oscillate with the same frequency, phase and amplitude?
(a) Frequency
The time-dependent part is:
cos(120πt)
Therefore, all oscillating points have the same angular frequency.
ω = 120π rad/s
Frequency:
ν = ω/2π
ν = 60 Hz
Hence, all points have the same frequency.
(b) Phase
All points between two consecutive nodes oscillate in the same phase.
Points in adjacent sections separated by a node oscillate in opposite phases.
Therefore, the phase is not the same for every point along the entire string.
(c) Amplitude
The amplitude is:
A(x) = 0.06 sin(2πx/3)
It depends on the position x.
Therefore, different points have different amplitudes.
(ii) Calculate amplitude at 0.375 m
Given:
x = 0.375 m
A(x) = 0.06 sin(2πx/3)
Substituting:
A = 0.06 sin(2π × 0.375/3)
A = 0.06 sin(π/4)
A = 0.06 × 0.707
A = 0.0424 m
Final Answer:
(i)
Frequency: Same for all oscillating points.
Phase: Same within each section between adjacent nodes; opposite in neighbouring sections.
Amplitude: Different at different points.
(ii) Amplitude = 0.0424 m or 4.24 cm.
Q14.13. Given below are some functions of x and t to represent the displacement 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√(x − vt)
(c) y = 3 sin(5x − 0.5t) + 4 cos(5x − 0.5t)
(d) y = cos x sin t + cos 2x sin 2t
Answer:
(a) y = 2 cos(3x) sin(10t)
This is the product of a function of position and a function of time.
It has the form:
y = f(x)g(t)
Therefore, it represents a stationary wave.
(b) y = 2√(x − vt)
The expression under the square root becomes negative for some values of x and t.
Therefore, the function is not defined over the entire real x-axis and cannot represent a physical wave over the entire line.
(c) y = 3 sin(5x − 0.5t) + 4 cos(5x − 0.5t)
Both terms contain the same combination:
5x − 0.5t
Their sum is also a function of this combination.
Therefore, it represents a travelling wave moving in the positive x-direction.
(d) y = cos x sin t + cos 2x sin 2t
Each term is a product of a function of position and a function of time.
Each term represents a stationary-wave component.
Therefore, it represents a superposition of stationary-wave components.
Final Answer:
(a) Stationary wave.
(b) None over the entire real x-axis.
(c) Travelling wave.
(d) Superposition of stationary-wave components.
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 × 10⁻² kg and its linear mass density is 4.0 × 10⁻² kg/m. Calculate (a) the speed of a transverse wave on the string and (b) the tension in the string.
Answer:
Given:
Frequency, ν = 45 Hz
Mass, m = 3.5 × 10⁻² kg
Linear mass density, μ = 4.0 × 10⁻² kg/m
Step 1: Calculate the length of the wire
μ = m/L
Therefore,
L = m/μ
L = (3.5 × 10⁻²)/(4.0 × 10⁻²)
L = 0.875 m
Step 2: Calculate the wavelength in the fundamental mode
For a string fixed at both ends in its fundamental mode:
λ = 2L
λ = 2 × 0.875
λ = 1.75 m
(a) Calculate wave speed
v = νλ
v = 45 × 1.75
v = 78.75 m/s
(b) Calculate tension
Formula:
v = √(T/μ)
Squaring:
T = μv²
T = 0.04 × 78.75²
T = 0.04 × 6201.56
T = 248.06 N
Final Answer:
(a) Speed = 78.75 m/s.
(b) Tension = approximately 248 N.
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. The edge effects may be neglected.
Answer:
Given:
Frequency, ν = 340 Hz
First resonance length:
L₁ = 25.5 cm
Second resonance length:
L₂ = 79.3 cm
For a tube closed at one end, two successive resonance lengths differ by λ/2.
Step 1: Calculate the difference between resonance lengths
L₂ − L₁ = λ/2
λ/2 = 79.3 − 25.5
λ/2 = 53.8 cm
Step 2: Calculate wavelength
λ = 2 × 53.8
λ = 107.6 cm
Converting into metres:
λ = 1.076 m
Step 3: Calculate speed of sound
Formula:
v = νλ
v = 340 × 1.076
v = 365.84 m/s
Final Answer:
Speed of sound = approximately 366 m/s.
Q14.16. A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is 2.53 kHz. What is the speed of sound in steel?
Answer:
Given:
Length of rod, L = 100 cm = 1 m
Frequency, ν = 2.53 kHz
ν = 2.53 × 10³ Hz
For a rod clamped at its middle in the fundamental mode:
The middle is a node.
Both free ends are antinodes.
Therefore, the length of the rod is half a wavelength.
L = λ/2
Hence,
λ = 2L
λ = 2 × 1
λ = 2 m
Wave speed:
v = νλ
v = 2.53 × 10³ × 2
v = 5.06 × 10³ m/s
Final Answer:
Speed of sound in steel = 5.06 × 10³ m/s or 5.06 km/s.
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? Take the speed of sound in air as 340 m/s.
Answer:
Given:
Length of pipe, L = 20 cm = 0.20 m
Frequency of source = 430 Hz
Speed of sound, v = 340 m/s
Part 1: Pipe closed at one end
The fundamental frequency of a closed pipe is:
ν₁ = v/4L
Substituting:
ν₁ = 340/(4 × 0.20)
ν₁ = 340/0.80
ν₁ = 425 Hz
The source frequency is 430 Hz, which is close to the fundamental frequency of 425 Hz.
Therefore, the fundamental mode (first harmonic) is resonantly excited approximately.
Part 2: Pipe open at both ends
The fundamental frequency of an open pipe is:
ν₁ = v/2L
ν₁ = 340/(2 × 0.20)
ν₁ = 340/0.40
ν₁ = 850 Hz
The higher harmonics are:
850 Hz, 1700 Hz, 2550 Hz, ...
The source frequency of 430 Hz does not match any of these frequencies.
Final Answer:
For the closed pipe, the fundamental mode is approximately excited.
For the pipe open at both ends, the 430 Hz source will not produce resonance.
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?
Answer:
Given:
Frequency of string A:
νA = 324 Hz
Initial beat frequency = 6 Hz
Final beat frequency = 3 Hz
Beat frequency formula:
νbeat = |νA − νB|
Step 1: Calculate the possible frequencies of B
6 = |324 − νB|
Therefore, two possibilities are:
νB = 324 + 6 = 330 Hz
or
νB = 324 − 6 = 318 Hz
Step 2: Identify the correct frequency
The tension of string A is reduced.
The frequency of a stretched string is proportional to the square root of its tension.
Therefore, reducing tension decreases the frequency of A.
The beat frequency also decreases from 6 Hz to 3 Hz.
This means the frequency of A moves closer to the frequency of B.
If B were 330 Hz, decreasing A below 324 Hz would increase the difference.
Therefore, B cannot have a frequency of 330 Hz.
Hence, the frequency of B is 318 Hz.
Final Answer:
Frequency of string B = 318 Hz.
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 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.
Answer:
(a) Why is a displacement node a pressure antinode?
A displacement node is a point where the particles of the medium have zero displacement at all times.
At this point, the neighbouring particles undergo maximum changes in compression and rarefaction.
Therefore, the pressure variation is maximum.
Hence, a displacement node corresponds to a pressure antinode.
Similarly, a displacement antinode corresponds to a pressure node.
(b) How can bats locate obstacles without eyes?
Bats use a method called echolocation.
They produce ultrasonic sound pulses.
These sound waves strike nearby obstacles.
The waves are reflected back as echoes.
Bats detect the returning echoes.
The time taken by the echo helps them estimate the distance of an obstacle.
The direction and characteristics of the reflected sound provide information about the obstacle.
Thus, bats can locate and identify obstacles using sound.
(c) Why do violin and sitar notes sound different even at the same frequency?
The fundamental frequencies of the two notes may be the same.
However, a violin and a sitar produce different combinations of overtones or harmonics.
The relative intensities of these harmonics are different.
This produces a different quality of sound, called timbre.
Therefore, we can distinguish between a violin note and a sitar note even when their fundamental frequencies are the same.
(d) Why can solids support both longitudinal and transverse waves, but gases support only longitudinal mechanical waves?
In longitudinal waves, particles vibrate parallel to the direction of propagation.
In transverse waves, particles vibrate perpendicular to the direction of propagation.
Solids possess both:
Bulk elasticity.
Shear elasticity.
Therefore, solids can support both longitudinal and transverse mechanical waves.
Gases do not possess a sustained shear restoring force.
Therefore, gases cannot support transverse mechanical waves.
However, they can support longitudinal sound waves because they can undergo compression and expansion.
(e) Why does a pulse get distorted in a dispersive medium?
A pulse consists of several wave components having different frequencies.
In a dispersive medium, the speed of a wave depends on its frequency.
Therefore, different frequency components travel at different speeds.
As they travel, they gradually separate from one another.
This changes the shape of the original pulse.
Hence, a pulse gets distorted during propagation through a dispersive medium.
IMPORTANT QUESTIONS AND ANSWERS
1. Derive the expression for the speed of a transverse wave on a stretched string.
Answer:
Consider a stretched string having:
T = Tension in the string
μ = Mass per unit length
The speed of a transverse wave on a stretched string depends on the tension and the linear mass density.
The expression is:
v = √(T/μ)
This means:
Wave speed increases when tension increases.
Wave speed decreases when linear mass density increases.
The speed does not depend on the amplitude or frequency of the wave for an ideal uniform string under constant tension.
2. Derive the general equation of a progressive wave.
Answer:
Consider a wave travelling along the positive x-direction.
Let:
A = Amplitude.
ω = Angular frequency.
k = Wave number.
φ = Initial phase.
At the origin, the particle executes SHM.
Its displacement is:
y(0, t) = A sin(ωt + φ)
A point at distance x from the origin has a phase difference:
kx = 2πx/λ
Therefore, the displacement of the particle at position x and time t is:
y(x, t) = A sin(ωt − kx + φ)
This is the equation of a progressive wave travelling in the positive x-direction.
For a wave travelling in the negative x-direction:
y(x, t) = A sin(ωt + kx + φ)
3. Derive the relation between wave speed, frequency and wavelength.
Answer:
Let a wave travel a distance equal to one wavelength λ in one time period T.
Wave speed is:
v = Distance/Time
Therefore,
v = λ/T
But:
ν = 1/T
Substituting:
v = νλ
This is the required relation.
4. Explain the principle of superposition of waves.
Answer:
According to the principle of superposition, when two or more waves travel through the same medium at the same time, the resultant displacement of a particle is equal to the algebraic sum of the displacements produced by the individual waves.
If two waves produce displacements y₁ and y₂, then:
y = y₁ + y₂
This principle explains:
Interference.
Formation of stationary waves.
Beats.
5. Explain the formation of stationary waves.
Answer:
Stationary waves are formed by the superposition of two waves having:
Equal amplitudes.
Equal frequencies.
Equal wavelengths.
Equal speeds.
Opposite directions of propagation.
Consider two waves:
y₁ = A sin(kx − ωt)
y₂ = A sin(kx + ωt)
According to the principle of superposition:
y = y₁ + y₂
y = A sin(kx − ωt) + A sin(kx + ωt)
Using the trigonometric identity:
sin(A − B) + sin(A + B) = 2 sin A cos B
Therefore,
y = 2A sin(kx) cos(ωt)
This is the equation of a stationary wave.
6. Differentiate between progressive waves and stationary waves.
Answer:
| Progressive Waves | Stationary Waves |
|---|---|
| The wave travels through the medium. | The wave pattern remains fixed in position. |
| Energy is transferred from one point to another. | There is no net energy transfer along the medium. |
| All particles have the same amplitude in an ideal uniform medium. | Amplitude varies from point to point. |
| Phase changes continuously with position. | Points within one segment between adjacent nodes oscillate in the same phase. |
| There are no fixed nodes and antinodes. | Fixed nodes and antinodes are formed. |
| Example: A travelling wave on a string. | Example: A vibrating string fixed at both ends. |
7. Derive the expression for the normal frequencies of a stretched string fixed at both ends.
Answer:
Consider a string of length L fixed at both ends.
When a stationary wave is formed, both ends must be nodes.
For the fundamental mode:
L = λ₁/2
Therefore,
λ₁ = 2L
Frequency:
ν₁ = v/λ₁
ν₁ = v/2L
For the second mode:
L = 2λ₂/2
Therefore,
λ₂ = L
ν₂ = v/L = 2v/2L
For the third mode:
L = 3λ₃/2
Therefore,
λ₃ = 2L/3
ν₃ = 3v/2L
In general:
λₙ = 2L/n
Therefore,
νₙ = nv/2L
where n = 1, 2, 3, ...
These are the normal frequencies of a string fixed at both ends.
8. Derive the expression for the frequencies of an open organ pipe.
Answer:
An open organ pipe has both ends open.
At each open end, a displacement antinode is formed.
For the fundamental mode:
L = λ₁/2
Therefore,
λ₁ = 2L
Frequency:
ν₁ = v/2L
For higher modes:
λₙ = 2L/n
Therefore,
νₙ = nv/2L
where n = 1, 2, 3, ...
An open pipe produces all harmonics.
9. Derive the expression for the frequencies of a closed organ pipe.
Answer:
A closed organ pipe has one end closed and the other end open.
At the closed end, a displacement node is formed.
At the open end, a displacement antinode is formed.
For the fundamental mode:
L = λ₁/4
Therefore,
λ₁ = 4L
Frequency:
ν₁ = v/4L
For the next allowed mode:
L = 3λ₃/4
Therefore,
λ₃ = 4L/3
Frequency:
ν₃ = 3v/4L
In general:
λₙ = 4L/(2n − 1)
Therefore,
νₙ = (2n − 1)v/4L
where n = 1, 2, 3, ...
A closed organ pipe produces only odd harmonics.
10. What are beats? Derive the expression for beat frequency.
Answer:
Beats are periodic variations in the loudness of sound produced by two sound waves of slightly different frequencies.
Let the two waves be:
y₁ = A cos(ω₁t)
y₂ = A cos(ω₂t)
According to the principle of superposition:
y = A cos(ω₁t) + A cos(ω₂t)
Using the identity:
cos A + cos B = 2 cos[(A − B)/2] cos[(A + B)/2]
Therefore,
y = 2A cos[(ω₁ − ω₂)t/2] cos[(ω₁ + ω₂)t/2]
The amplitude changes with time according to:
Amplitude = 2A cos[(ω₁ − ω₂)t/2]
The number of loudness maxima per second is:
νbeat = |ν₁ − ν₂|
This is the required expression for beat frequency.
11. Explain the difference between Newton's formula and Laplace's correction for the speed of sound in a gas.
Answer:
Newton's Formula:
Newton assumed that the compressions and rarefactions in a sound wave take place isothermally.
The bulk modulus of a gas under isothermal conditions is:
B = P
Therefore, speed of sound:
v = √(P/ρ)
This gives a value lower than the experimentally observed speed.
Laplace's Correction:
Laplace explained that sound travels through a gas so rapidly that there is very little time for heat exchange.
Therefore, compressions and rarefactions are approximately adiabatic.
The bulk modulus is:
B = γP
Therefore, the corrected speed of sound is:
v = √(γP/ρ)
where γ is the ratio of specific heats.
Laplace's formula gives a value much closer to the experimental speed.
12. Explain the Doppler effect with an example.
Answer:
The Doppler effect is the apparent change in frequency of a wave due to relative motion between the source and the observer.
When the source approaches the observer:
The observed frequency increases.
Therefore, the sound appears to have a higher pitch.
When the source moves away from the observer:
The observed frequency decreases.
Therefore, the sound appears to have a lower pitch.
Example:
The siren of an ambulance sounds higher as it approaches a person and lower as it moves away.
IMPORTANT FORMULAE – CHAPTER 14
Wave speed:
v = νλ
Frequency:
ν = 1/T
Time period:
T = 1/ν
Angular frequency:
ω = 2πν = 2π/T
Wave number:
k = 2π/λ
Speed of wave:
v = ω/k
Progressive wave moving in positive x-direction:
y = A sin(ωt − kx + φ)
Progressive wave moving in negative x-direction:
y = A sin(ωt + kx + φ)
Speed of transverse wave on a string:
v = √(T/μ)
Linear mass density:
μ = m/L
Speed of sound in a fluid:
v = √(B/ρ)
Speed of sound in a solid rod:
v = √(Y/ρ)
Newton's formula for speed of sound:
v = √(P/ρ)
Laplace's formula for speed of sound:
v = √(γP/ρ)
Speed of sound in an ideal gas:
v = √(γRT/M)
Phase difference:
Δφ = 2πΔx/λ
Stationary wave equation:
y = 2A sin(kx) cos(ωt)
Distance between consecutive nodes:
λ/2
Distance between consecutive antinodes:
λ/2
Distance between a node and its nearest antinode:
λ/4
Frequency of a string fixed at both ends:
νₙ = nv/2L
Frequency of an open organ pipe:
νₙ = nv/2L
Frequency of a closed organ pipe:
νₙ = (2n − 1)v/4L
Beat frequency:
νbeat = |ν₁ − ν₂|
Wavelength:
λ = v/ν
IMPORTANT SI UNITS
| Physical Quantity | SI Unit |
|---|---|
| Displacement | metre (m) |
| Amplitude | metre (m) |
| Wavelength | metre (m) |
| Frequency | hertz (Hz) |
| Time period | second (s) |
| Wave speed | metre per second (m/s) |
| Angular frequency | radian per second (rad/s) |
| Wave number | radian per metre (rad/m) |
| Tension | newton (N) |
| Linear mass density | kg/m |
| Pressure | pascal (Pa) |
| Density | kg/m³ |
| Bulk modulus | pascal (Pa) |
IMPORTANT POINTS TO REMEMBER
A wave transfers energy without transferring matter as a whole.
Mechanical waves require a material medium.
Transverse waves involve particle vibration perpendicular to wave propagation.
Longitudinal waves involve particle vibration parallel to wave propagation.
Sound waves in air are longitudinal waves.
Wave speed is given by v = νλ.
Frequency is determined by the source and remains unchanged when a wave enters another medium.
Wavelength changes when the wave speed changes in a new medium.
The principle of superposition explains interference and stationary waves.
Constructive interference occurs when waves meet in the same phase.
Destructive interference occurs when waves meet in opposite phases.
Stationary waves contain fixed nodes and antinodes.
There is no net energy transfer along an ideal stationary wave.
A string fixed at both ends produces all harmonics.
An open organ pipe produces all harmonics.
A closed organ pipe produces only odd harmonics.
Beats are produced by two sound waves having slightly different frequencies.
Beat frequency is the absolute difference between the two frequencies.
The speed of sound in air increases with temperature.
The speed of sound in moist air is greater than in dry air at the same temperature and pressure.
Laplace corrected Newton's formula by considering sound propagation to be approximately adiabatic.
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