Waves

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Summary

Chapter 14: Waves

Summary

Mechanical waves exist in material media and follow Newton's Laws.
Transverse Waves: Particles oscillate perpendicular to wave direction.
Longitudinal Waves: Particles oscillate along the direction of wave propagation.
Progressive Wave: Moves from one point in the medium to another.
Displacement Relation: For a sinusoidal wave:
$y(x, t) = a \sin(kx - \omega t + \Phi)$ $y (x, t) = a sin (k x - ω t + Φ)$
where:
- a = amplitude
- k = angular wave number
- ω = angular frequency
- Φ = phase constant
Wavelength (λ): Distance between two consecutive points of the same phase.
Period (T): Time for one complete oscillation, related to angular frequency (ω) by:
$T = \frac{2\pi}{\omega}$
Frequency (ν): Defined as 1/T, related to angular frequency by:
$ν = \frac{\omega}{2\pi}$
Wave Speed (v): Given by:
$v = \frac{A}{kT}$
Speed of Transverse Wave on String:
$v_{TH} = \sqrt{\frac{T}{µ}}$
where T = tension, µ = linear mass density.
Speed of Sound in Fluids:
$v = \sqrt{\frac{B}{ρ}}$
where B = bulk modulus, ρ = density.

Key Formulas/Definitions

Physical Quantity	Symbol	Dimensions	Unit	Remarks
Wavelength	λ	[L]	m	Distance between two consecutive points with the same phase.
Propagation Constant	k	[L⁻¹]	m⁻¹	k = 2π/λ
Wave Speed	v	[LT⁻¹]	m/s	Speed of wave propagation.
Beat Frequency	νₘₑₐₜ	[T⁻¹]	s⁻¹	Difference of two close frequencies of superposing waves.

Points to Ponder

A wave does not involve the motion of matter as a whole in a medium.
Energy, not matter, is transferred in a wave.
Mechanical waves transfer energy through elastic forces between oscillating parts of the medium.
Transverse waves require shear modulus; longitudinal waves require bulk modulus.
In harmonic progressive waves, all particles have the same amplitude but different phases.
The speed of a mechanical wave depends on the medium's properties, not the source's velocity.

Learning Objectives

Understand the properties and behavior of waves in different media.
Differentiate between transverse and longitudinal waves.
Analyze the displacement relation in a progressive wave.
Calculate the speed of a travelling wave using relevant formulas.
Apply the principle of superposition of waves to solve problems.
Explain the phenomenon of wave reflection.
Investigate the concept of beats in wave motion.

Detailed Notes

Chapter 14: Waves

14.1 Introduction

Study of waves in a material medium.
Waves transport energy without the physical transfer of matter.

14.2 Transverse and Longitudinal Waves

Transverse Waves: Particles oscillate perpendicular to wave direction.
Longitudinal Waves: Particles oscillate along the direction of wave propagation.

14.3 Displacement Relation in a Progressive Wave

Displacement in a sinusoidal wave:

$y(x, t) = a \, sin(kx - wt + \Phi)$
- Where:
  - a = amplitude
  - k = angular wave number
  - w = angular frequency
  - Φ = phase constant

14.4 The Speed of a Travelling Wave

Speed of a progressive wave:

$v = \frac{A}{T}$
- Speed of transverse wave on a string:
$v_{TH} = \sqrt{\frac{T}{\mu}}$
- Where:
  - T = tension
  - μ = linear mass density

14.5 The Principle of Superposition of Waves

Superposition principle applies to waves, allowing for interference patterns.

14.6 Reflection of Waves

Waves can reflect off boundaries, changing direction without loss of energy.

14.7 Beats

Beats occur when two waves of slightly different frequencies interfere.
Beat frequency:

$V_{beat} = |f_1 - f_2|$

Summary

Mechanical waves exist in material media and follow Newton's Laws.
Transverse waves oscillate perpendicular to propagation direction.
Longitudinal waves oscillate along the propagation direction.
Progressive waves move from one point to another.
Wavelength is the distance between consecutive points of the same phase.
Period T is the time for one complete oscillation.
Frequency v is the inverse of period T.
Speed of sound in a medium depends on its properties.

Points to Ponder

Waves transfer energy, not matter.
Transverse waves require shear modulus; longitudinal waves require bulk modulus.
Speed of mechanical waves depends on medium properties, not source velocity.

Exercises

Various problems related to wave speed, frequency, and tension in strings.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Wave Motion: Students often confuse the motion of waves with the motion of matter. Remember, waves transport energy, not matter.
Confusing Transverse and Longitudinal Waves: Ensure you understand that in transverse waves, particles move perpendicular to wave direction, while in longitudinal waves, they move parallel.
Ignoring the Medium's Properties: The speed of sound and other waves depends on the medium's properties (like density and elasticity). Don't assume it is constant across different media.
Neglecting Phase Relationships: When dealing with interference, be careful with phase differences. Constructive and destructive interference depend on the phase relationship between waves.

Exam Tips

Understand Key Definitions: Be clear on definitions like wavelength, frequency, and wave speed. These are often tested directly.
Practice Wave Equations: Familiarize yourself with the equations for wave speed, frequency, and wavelength. Being able to manipulate these equations is crucial.
Visualize Problems: Draw diagrams for wave interactions, such as reflection and interference. Visual aids can help clarify complex concepts.
Review Examples: Go through example problems, especially those involving calculations of wave speed in different media and the effects of tension on wave speed in strings.
Check Units: Always ensure your units are consistent when performing calculations, especially in formulas involving speed, frequency, and wavelength.

Practice & Assessment

Multiple Choice Questions

0.5 m

1 m

2 m

4 m

Correct Answer: B

Solution:

The wave equation is of the form

y(x, t) = a \sin(kx - \omega t)

, where

k = 4\pi

. The wavelength

\lambda

is given by

\lambda = \frac{2\pi}{k} = \frac{2\pi}{4\pi} = 0.5

v = \sqrt{\frac{B}{\rho}}

v = \frac{B}{\rho}

v = B \rho

v = \sqrt{B \rho}

Correct Answer: A

Solution:

The speed of sound wave in a fluid having bulk modulus

B

and density

\rho

is given by

v = \sqrt{\frac{B}{\rho}}

20 m/s

40 m/s

80 m/s

100 m/s

Correct Answer: B

Solution:

The speed of a transverse wave on a string is given by

v = \sqrt{\frac{T}{\mu}}

, where

T

is the tension and

\mu

is the linear mass density. Here,

\mu = \frac{2.50 \text{ kg}}{20.0 \text{ m}} = 0.125 \text{ kg/m}

. Thus,

v = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = 40 \text{ m/s}.

v

2v

\sqrt{2}v

4v

Correct Answer: B

Solution:

The speed of a wave on a string is given by

v = \sqrt{\frac{T}{\mu}}

. If the tension is increased to

4T

, the new speed

v' = \sqrt{\frac{4T}{\mu}} = 2v

Zero

Equal to the amplitude of one wave

Twice the amplitude of one wave

Half the amplitude of one wave

Correct Answer: A

Solution:

When two waves are out of phase by

\pi

, they interfere destructively, resulting in a net amplitude of zero. This is due to the principle of superposition.

Positive x-direction

Negative x-direction

Positive y-direction

Negative y-direction

Correct Answer: B

Solution:

The wave equation is of the form

y(x, t) = a \cos(kx + \omega t)

. The positive sign in

kx + \omega t

indicates that the wave is traveling in the negative x-direction.

Particles oscillate parallel to the direction of wave propagation.

Particles oscillate perpendicular to the direction of wave propagation.

Particles move in a circular motion.

Particles do not move at all.

Correct Answer: B

Solution:

In a transverse wave, particles of the medium oscillate perpendicular to the direction of wave propagation.

\pi

\frac{\pi}{2}

2\pi

\frac{3\pi}{2}

Correct Answer: C

Solution:

For constructive interference, the phase difference between the waves is an integral multiple of

2\pi

It determines the frequency of the wave.

It determines the amplitude of the wave.

It determines the initial phase angle of the wave.

It determines the speed of the wave.

Correct Answer: C

Solution:

The phase constant, denoted as

\Phi

, determines the initial phase angle of the wave.

150 m/s

450 m/s

300 m/s

450\sqrt{3} m/s

Correct Answer: C

Solution:

The speed of a wave on a string is given by

v = \sqrt{\frac{T}{\mu}}

, where

T

is the tension and

\mu

is the linear mass density. If the tension is increased by a factor of 9, the new speed

v' = \sqrt{9}v = 3v = 3 \times 150 = 300 \text{ m/s}

The wavelength in air is longer than in water.

The wavelength in water is longer than in air.

The wavelengths in both media are equal.

The wavelength in air is zero.

Correct Answer: B

Solution:

The speed of a wave

v

is given by

v = f\lambda

, where

f

is the frequency and

\lambda

is the wavelength. Since the frequency is constant, the wavelength is directly proportional to the speed. Thus, the wavelength in water is longer than in air.

A wave that remains stationary.

A wave that moves from one point of medium to another.

A wave that only oscillates at the source.

A wave that decreases in amplitude over time.

Correct Answer: B

Solution:

A progressive wave is one that moves from one point of the medium to another.

It can exist in a vacuum.

It requires a material medium to propagate.

It does not transfer energy.

It travels at the speed of light.

Correct Answer: B

Solution:

Mechanical waves require a material medium to propagate, as they are governed by Newton's Laws.

The speed decreases.

The speed remains the same.

The speed increases.

The speed becomes zero.

Correct Answer: C

Solution:

The speed of sound in air is given by

v = \sqrt{\frac{\gamma P}{\rho}}

. As temperature increases, the density

\rho

decreases, leading to an increase in the speed of sound.

4 Hz

256 Hz

258 Hz

260 Hz

Correct Answer: A

Solution:

The beat frequency is the absolute difference between the two frequencies:

f_{\text{beat}} = |260 - 256| = 4

Hz.

It changes the frequency of the wave.

It changes the speed of the wave.

It shifts the wave in space.

It changes the amplitude of the wave.

Correct Answer: C

Solution:

The phase constant

\Phi

shifts the wave along the x-axis without affecting its frequency, speed, or amplitude. It represents the initial phase of the wave at

x = 0

and

t = 0

100 Hz

200 Hz

300 Hz

400 Hz

Correct Answer: A

Solution:

For a closed pipe, the fundamental frequency is given by

f = \frac{v}{4L} = \frac{340}{4 \times 0.85} = 100

Hz.

v = T

v = 2\pi T

v = 1/T

v = T^2

Correct Answer: C

Solution:

The frequency

v

of a wave is defined as the reciprocal of the period

T

, i.e.,

v = 1/T

They are the same point.

A displacement node is a pressure antinode.

A displacement node is a pressure node.

They do not relate to each other.

Correct Answer: B

Solution:

In a sound wave, a displacement node is a pressure antinode and vice versa, meaning that where the displacement is minimal, the pressure variation is maximal.

1500 N

2000 N

2500 N

3000 N

Correct Answer: C

Solution:

The speed of a transverse wave on a string is

v = \sqrt{\frac{T}{\mu}}

. Here,

v = 343 \text{ m/s}

\mu = \frac{1.50}{10.0} = 0.15 \text{ kg/m}

. Solving for

T

343 = \sqrt{\frac{T}{0.15}} \Rightarrow T = 343^2 \times 0.15 = 2500 \text{ N}

There are three nodes and two antinodes.

There are two nodes and three antinodes.

There are four nodes and three antinodes.

There are three nodes and four antinodes.

Correct Answer: C

Solution:

In the third harmonic of a string fixed at both ends, there are four nodes (including the endpoints) and three antinodes.

Zero

Double the amplitude of one wave

Half the amplitude of one wave

Same as the amplitude of one wave

Correct Answer: A

Solution:

When two waves are out of phase by

\pi

, they interfere destructively, resulting in zero amplitude:

A(\Phi) = 2a \cos\left(\frac{\Phi}{2}\right) = 2a \cos\left(\frac{\pi}{2}\right) = 0

0.5 m

1.0 m

2.0 m

0.25 m

Correct Answer: B

Solution:

In a standing wave, the distance between two consecutive nodes is half the wavelength. Therefore, the wavelength

\lambda = 2 \times 0.5 = 1.0 \text{ m}

f

2f

\frac{f}{2}

\frac{3f}{2}

Correct Answer: B

Solution:

For a pipe open at one end, the fundamental frequency is

v/4L

. When open at both ends, the fundamental frequency becomes

v/2L

, which is twice the original frequency.

They can only propagate in solids.

They require a medium with shear modulus of elasticity.

Particles oscillate along the direction of wave propagation.

They cannot travel through gases.

Correct Answer: C

Solution:

In longitudinal waves, particles of the medium oscillate along the direction of wave propagation.

\frac{v}{2L}

\frac{v}{L}

\frac{2v}{L}

\frac{v}{4L}

Correct Answer: A

Solution:

The fundamental frequency of a string fixed at both ends is given by

f = \frac{v}{2L}

, where

v

is the speed of the wave and

L

is the length of the string.

The displacement of a medium is the product of the displacements due to each wave.

The displacement of a medium is the algebraic sum of the displacements due to each wave.

The displacement of a medium is the difference between the displacements due to each wave.

The displacement of a medium is unaffected by the presence of other waves.

Correct Answer: B

Solution:

The principle of superposition states that the displacement of any element of the medium is the algebraic sum of the displacements due to each wave.

0.28 m

0.57 m

0.85 m

1.13 m

Correct Answer: A

Solution:

For a pipe closed at one end, the fundamental frequency is given by

f = \frac{v}{4L}

. Solving for

L

, we have

L = \frac{v}{4f} = \frac{340}{4 \times 300} = 0.28

0.5 seconds

0.7 seconds

0.9 seconds

1.0 seconds

Correct Answer: B

Solution:

The speed of a transverse wave on a string is given by

v = \sqrt{\frac{T}{\mu}}

, where

T

is the tension and

\mu

is the linear mass density. Here,

T = 180 \text{ N}

and

\mu = \frac{3.00}{15.0} = 0.2 \text{ kg/m}

. Thus,

v = \sqrt{\frac{180}{0.2}} = 30 \text{ m/s}

. The time taken to travel 15.0 m is

t = \frac{15.0}{30} = 0.5 \text{ s}

1667 Hz

3333 Hz

2500 Hz

5000 Hz

Correct Answer: A

Solution:

For a rod clamped at its midpoint, the fundamental frequency is given by

f = \frac{v}{2L}

. Therefore,

f = \frac{5000}{2 \times 1.5} = 1667 \text{ Hz}

Matter

Energy

Both matter and energy

Neither matter nor energy

Correct Answer: B

Solution:

In a mechanical wave, energy is transferred from one point to another, not matter.

50 m/s

100 m/s

200 m/s

400 m/s

Correct Answer: B

Solution:

The wave equation is of the form

y(x, t) = a \sin(kx - \omega t)

. Here,

k = 4\pi

and

\omega = 200\pi

. The speed

v

of the wave is given by

v = \frac{\omega}{k} = \frac{200\pi}{4\pi} = 50

m/s.

Zero

Equal to the amplitude of one wave

Twice the amplitude of one wave

Four times the amplitude of one wave

Correct Answer: C

Solution:

Constructive interference occurs when two waves are in phase, resulting in a superposed wave with an amplitude that is the sum of the individual amplitudes. Thus, the resulting amplitude is twice the amplitude of one wave.

They can exist in a vacuum.

They require a material medium to propagate.

They do not transfer energy.

They are not governed by Newton's Laws.

Correct Answer: B

Solution:

Mechanical waves require a material medium to propagate and are governed by Newton's Laws.

kx - wt + \Phi

a \sin(kx - wt + \Phi)

kx + wt

a

Correct Answer: A

Solution:

The phase of the wave is the term inside the sine function, which is

kx - wt + \Phi

2 Hz

4 Hz

6 Hz

8 Hz

Correct Answer: B

Solution:

The beat frequency is the absolute difference between the two frequencies, which is 4 Hz.

The speed of sound decreases.

The speed of sound remains constant.

The speed of sound increases.

The speed of sound becomes zero.

Correct Answer: C

Solution:

The speed of sound in air increases with temperature because the speed is proportional to the square root of the temperature.

First harmonic

Second harmonic

Third harmonic

Fourth harmonic

Correct Answer: C

Solution:

For a pipe closed at one end, the resonant frequencies are given by

f_n = \frac{(2n-1)v}{4L}

, where

n

is the harmonic number,

v

is the speed of sound, and

L

is the length of the pipe. Solving for

n

with

f_n = 430

Hz,

v = 340

m/s, and

L = 0.2

m, we find

430 = \frac{(2n-1) \times 340}{4 \times 0.2}

. Solving gives

n = 3

, which corresponds to the third harmonic.

4 m

2 m

8 m

6 m

Correct Answer: A

Solution:

The speed of a wave is given by the product of its frequency and wavelength:

v = f \lambda

. Rearranging for wavelength, we have

\lambda = \frac{v}{f} = \frac{200 \text{ m/s}}{50 \text{ Hz}} = 4 \text{ m}

Particles oscillate parallel to the direction of wave propagation.

Particles oscillate perpendicular to the direction of wave propagation.

Particles do not oscillate at all.

Particles oscillate in a circular motion.

Correct Answer: B

Solution:

In transverse waves, the particles of the medium oscillate perpendicular to the direction of wave propagation.

90 Hz

135 Hz

270 Hz

360 Hz

Correct Answer: B

Solution:

The frequency of a string vibrating in its nth harmonic is given by

f_n = \frac{nv}{2L}

, where

n

is the harmonic number,

v

is the speed of the wave, and

L

is the length of the string. For the third harmonic (

n = 3

f_3 = \frac{3 \times 180}{2 \times 2.0} = 135 \text{ Hz}

A traveling wave with increased amplitude.

A standing wave with nodes and antinodes.

A wave with decreased amplitude traveling in one direction.

Complete cancellation, resulting in no wave.

Correct Answer: B

Solution:

When two identical waves travel in opposite directions, they superpose to form a standing wave characterized by nodes (points of zero displacement) and antinodes (points of maximum displacement).

The displacement of any element of the medium is the product of the displacements due to each wave.

The displacement of any element of the medium is the sum of the displacements due to each wave.

The displacement of any element of the medium is the difference of the displacements due to each wave.

The displacement of any element of the medium is the average of the displacements due to each wave.

Correct Answer: B

Solution:

The principle of superposition states that the displacement of any element of the medium is the algebraic sum of the displacements due to each wave.

Parallel to the direction of wave propagation

Perpendicular to the direction of wave propagation

In a circular motion

Do not oscillate

Correct Answer: B

Solution:

In transverse waves, the particles of the medium oscillate perpendicular to the direction of wave propagation.

2.5 m

3.0 m

2.0 m

1.5 m

Correct Answer: A

Solution:

The speed of a wave is given by the product of its frequency and wavelength, i.e.,

v = f\lambda

. Here,

v = 150 \text{ m/s}

and

f = 60 \text{ Hz}

. Thus,

\lambda = \frac{v}{f} = \frac{150}{60} = 2.5 \text{ m}

\lambda

\frac{\lambda}{4}

\frac{\lambda}{2}

2\lambda

Correct Answer: C

Solution:

In a standing wave, the distance between two consecutive nodes or antinodes is

\frac{\lambda}{2}

, where

\lambda

is the wavelength.

60 m/s

120 m/s

240 m/s

480 m/s

Correct Answer: C

Solution:

The speed of a wave on a string is given by

v = \sqrt{\frac{T}{\mu}}

. If the tension is quadrupled, the new speed is

v' = \sqrt{4}v = 2v = 240

m/s.

v = \sqrt{T/\mu}

v = T \cdot \mu

v = T/\mu

v = \mu/T

Correct Answer: A

Solution:

The speed of a transverse wave on a string is given by

v = \sqrt{T/\mu}

, where

T

is the tension and

\mu

is the linear mass density.

Transverse waves oscillate perpendicular to the direction of wave propagation, while longitudinal waves oscillate parallel.

Transverse waves oscillate parallel to the direction of wave propagation, while longitudinal waves oscillate perpendicular.

Both transverse and longitudinal waves oscillate perpendicular to the direction of wave propagation.

Both transverse and longitudinal waves oscillate parallel to the direction of wave propagation.

Correct Answer: A

Solution:

Transverse waves involve oscillations perpendicular to the direction of wave propagation, whereas longitudinal waves involve oscillations parallel to the direction of wave propagation.

318 Hz

321 Hz

327 Hz

330 Hz

Correct Answer: C

Solution:

Initially, the beat frequency is 6 Hz, so

|f_B - 324| = 6

. Thus,

f_B

could be 330 Hz or 318 Hz. After reducing the tension in A, the beat frequency becomes 3 Hz, indicating

|f_B - f'_A| = 3

. Since reducing tension decreases frequency,

f'_A < 324

. Hence,

f_B

must be 327 Hz to satisfy the new condition.

\sqrt{2}A

Correct Answer: C

Solution:

When two waves with the same amplitude

A

and a phase difference

\Phi = \frac{\pi}{2}

interfere, the resultant amplitude

A_r

is given by

A_r = \sqrt{A^2 + A^2 + 2A^2\cos(\Phi)} = \sqrt{2A^2} = \sqrt{2}A

a

k

w

x

Correct Answer: A

Solution:

The amplitude of the wave is represented by

a

in the equation

y(x, t) = a \sin(kx - wt)

Wavelength

Amplitude

Frequency

Phase constant

Correct Answer: B

Solution:

The term

a

represents the amplitude of the wave, which is the maximum displacement of the particles from their equilibrium position.

0.332 m

0.664 m

1.328 m

0.166 m

Correct Answer: A

Solution:

For a tube closed at one end, the fundamental frequency is given by

f = \frac{v}{4L}

. Solving for

L

, we have

L = \frac{v}{4f} = \frac{340}{4 \times 256} = 0.332 \text{ m}

Particles of the medium oscillate perpendicular to the direction of wave propagation.

Particles of the medium oscillate along the direction of wave propagation.

The wave travels in a circular motion.

The wave does not require a medium to propagate.

Correct Answer: B

Solution:

In longitudinal waves, particles of the medium oscillate along the direction of wave propagation.

T = \frac{2\pi}{\omega}

T = 2\pi \omega

T = \omega

T = \frac{1}{\omega}

Correct Answer: A

Solution:

The period

T

of oscillation of a wave is related to the angular frequency

\omega

T = \frac{2\pi}{\omega}

y(x, t) = a \sin(kx - wt + \Phi)

y(x, t) = a \cos(kx + wt + \Phi)

y(x, t) = a \sin(kx + wt + \Phi)

y(x, t) = a \cos(kx - wt + \Phi)

Correct Answer: A

Solution:

The displacement in a sinusoidal wave propagating in the positive X direction is given by

y(x, t) = a \sin(kx - wt + \Phi)

Displacement nodes are points of zero pressure variation.

Displacement nodes are points of maximum pressure variation.

Displacement antinodes are points of zero pressure variation.

Displacement antinodes are points of maximum pressure variation.

Correct Answer: B

Solution:

In a sound wave, displacement nodes (points of zero displacement) correspond to points of maximum pressure variation (pressure antinodes) because the compression and rarefaction of the medium are greatest where the displacement is zero.

50 Hz

100 Hz

150 Hz

200 Hz

Correct Answer: B

Solution:

For the second harmonic, the wavelength is

\lambda = \frac{2L}{2} = L = 1.5

m. The frequency is given by

f = \frac{v}{\lambda} = \frac{150}{1.5} = 100

Hz.

It is absorbed completely.

It passes through without any change.

It reflects with a phase reversal.

It reflects without any phase change.

Correct Answer: C

Solution:

When a wave meets a rigid boundary, it reflects with a phase reversal.

100 m/s

50 m/s

10 m/s

30 m/s

Correct Answer: A

Solution:

The speed of a transverse wave on a string is given by

v = \sqrt{\frac{T}{\mu}}

, where

T

is the tension and

\mu

is the linear mass density. Here,

\mu = \frac{3.00}{30.0} = 0.1

kg/m. Thus,

v = \sqrt{\frac{300}{0.1}} = 100

m/s.

The speed of the wave.

The maximum displacement of the medium's constituents from their equilibrium position.

The frequency of the wave.

The wavelength of the wave.

Correct Answer: B

Solution:

The amplitude of a wave represents the maximum displacement of the medium's constituents from their equilibrium position.

Particles oscillate perpendicular to the direction of wave propagation.

Particles oscillate parallel to the direction of wave propagation.

Particles move in a circular motion.

Particles remain stationary.

Correct Answer: A

Solution:

In transverse waves, the particles of the medium oscillate perpendicular to the direction of wave propagation.

2530 m/s

5060 m/s

1265 m/s

6325 m/s

Correct Answer: B

Solution:

For a rod clamped at its middle, the fundamental frequency is given by

f = \frac{v}{2L}

, where

v

is the speed of sound and

L

is the length of the rod. Rearranging gives

v = 2fL = 2 \times 2530 \times 0.5 = 5060 \text{ m/s}.

2 m

1 m

0.5 m

0.2 m

Correct Answer: A

Solution:

The wavelength

\lambda

of a wave is given by

\lambda = \frac{v}{f}

, where

v

is the speed of the wave and

f

is the frequency. Substituting the given values,

\lambda = \frac{150}{75} = 2

50 m/s

100 m/s

200 m/s

400 m/s

Correct Answer: B

Solution:

For a string fixed at both ends, the fundamental frequency is given by

f = \frac{v}{2L}

. Here,

f = 50

Hz and

L = 2

m. Solving for

v

, we have

v = 2Lf = 2 \times 2 \times 50 = 200

m/s.

0.5 s

1.0 s

1.5 s

2.0 s

Correct Answer: B

Solution:

The speed of a transverse wave on a string is given by

v = \sqrt{\frac{T}{\mu}}

, where

T

is the tension and

\mu

is the linear mass density.

\mu = \frac{4.00}{20.0} = 0.2

kg/m. Thus,

v = \sqrt{\frac{160}{0.2}} = 28.28

m/s. The time taken

t = \frac{20.0}{28.28} \approx 0.71

The wavelength is halved.

The wavelength is doubled.

The wavelength remains unchanged.

The wavelength becomes zero.

Correct Answer: A

Solution:

The wave number

k

is related to the wavelength

\lambda

by the equation

k = \frac{2\pi}{\lambda}

. If

k

is doubled, then

\lambda

must be halved to maintain the equation.

True or False

Correct Answer: False

Solution:

Transverse waves require a medium with shear modulus of elasticity, which gases do not possess.

Correct Answer: True

Solution:

Waves transport energy and the pattern of disturbance propagates, but there is no physical transfer of matter.

Correct Answer: True

Solution:

The speed of sound in air is independent of pressure, as explained by the formula involving the bulk modulus and density.

Correct Answer: True

Solution:

The speed of sound in a fluid is given by the formula

v = \sqrt{\frac{B}{\rho}}

, where

B

is the bulk modulus and

Solution:

The speed of sound in air increases with temperature because the speed is proportional to the square root of the temperature.

Correct Answer: True

Solution:

In a standing wave, nodes are points of zero displacement, and the distance between two consecutive nodes is half the wavelength,

\frac{\lambda}{2}

Correct Answer: True

Solution:

In sound waves, displacement nodes correspond to pressure antinodes and vice versa.

Correct Answer: False

Solution:

Transverse waves require a medium with shear modulus of elasticity, which is not present in liquids and gases. Therefore, they can only propagate in solids.

Correct Answer: True

Solution:

Mechanical waves transfer energy due to the coupling through elastic forces between neighboring oscillating parts of the medium.

Correct Answer: False

Solution:

In a progressive wave, particles have the same amplitude but different phases at a given instant of time.

Correct Answer: True

Solution:

This equation describes the displacement of a sinusoidal wave propagating in the positive X direction, where

a

is the amplitude,

k

is the angular wave number,

w

is the angular frequency, and

Solution:

The speed of a transverse wave on a string is given by the formula

v = \sqrt{\frac{T}{\mu}}

, where

T

is the tension and

Solution:

Solution:

The speed of sound in a gas is independent of pressure, as it depends on the bulk modulus and density of the gas.

Correct Answer: True

Solution:

When a wave pulse meets a rigid boundary, it reflects and inverts.

Waves

AI Learning Assistant

Summary

Chapter 14: Waves

Summary

Key Formulas/Definitions

Points to Ponder

Learning Objectives

Detailed Notes

Chapter 14: Waves

14.1 Introduction

14.2 Transverse and Longitudinal Waves

14.3 Displacement Relation in a Progressive Wave

14.4 The Speed of a Travelling Wave

14.5 The Principle of Superposition of Waves

14.6 Reflection of Waves

14.7 Beats

Summary

Points to Ponder

Exercises

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Exam Tips

Practice & Assessment

Multiple Choice Questions

Q1.A progressive wave is described by the equation y(x,t)=0.1sin⁡(4πx−100πt)y(x, t) = 0.1 \sin(4\pi x - 100\pi t)y(x,t)=0.1sin(4πx−100πt), where xxx and yyy are in meters and ttt is in seconds. What is the wavelength of this wave?

Correct Answer: B

Q2.What is the speed of sound in a fluid with bulk modulus BBB and density ρ\rhoρ?

Correct Answer: A

Q3.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 wave is generated on this string, what is the speed of the wave?

Correct Answer: B

Q4.A wave travels along a string with a speed vvv when the tension in the string is TTT and the linear mass density is μ\muμ. If the tension is increased to 4T4T4T, what will be the new speed of the wave?

Correct Answer: B

Q5.Two waves of equal amplitude and frequency travel in the same direction. If they have a phase difference of π\piπ, what is the resultant amplitude?

Correct Answer: A

Q6.A wave on a string is described by the equation y(x,t)=0.1cos⁡(5x+20t)y(x, t) = 0.1 \cos(5x + 20t)y(x,t)=0.1cos(5x+20t). What is the direction of wave propagation?

Correct Answer: B

Q7.What is the nature of the motion of particles in a transverse wave?

Correct Answer: B

Q8.In a medium, two waves with the same frequency and amplitude interfere constructively. What is the phase difference between them?

Correct Answer: C

Q9.What is the phase constant in the equation of a sinusoidal wave?

Correct Answer: C

Q10.A wave travels along a string with a speed of 150 m/s. If the tension in the string is increased by a factor of 9, what will be the new speed of the wave?

Correct Answer: C

Correct Answer: B

Q12.Which of the following best describes a progressive wave?

Correct Answer: B

Q13.What is the primary characteristic of a mechanical wave?

Correct Answer: B

Q14.A sound wave travels through air with a speed vvv. If the temperature of the air increases, what happens to the speed of sound?

Correct Answer: C

Q15.Two sound waves of frequencies 256 Hz and 260 Hz are played together. What is the beat frequency observed?

Correct Answer: A

Q16.A wave traveling along a string is described by the equation y(x,t)=Asin⁡(kx−ωt)y(x, t) = A \sin(kx - \omega t)y(x,t)=Asin(kx−ωt). If the phase constant Φ\PhiΦ is added to the equation, what effect does it have on the wave?

Correct Answer: C

Q17.In a closed pipe resonating at its fundamental frequency, the length of the pipe is 0.85 m. If the speed of sound is 340 m/s, what is the frequency of the sound wave?

Correct Answer: A

Q18.What is the relationship between frequency vvv and period TTT of a wave?

Correct Answer: C

Q19.In a sound wave, what is the relationship between a displacement node and a pressure antinode?

Correct Answer: B

Q20.A steel wire has a length of 10.0 m and a mass of 1.50 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 air at 20°C, which is 343 m/s?

Correct Answer: C

Q21.Consider a string of length LLL fixed at both ends. If the string vibrates in its third harmonic, which of the following statements is true about the nodes and antinodes?

Correct Answer: C

Q22.Two waves of the same frequency and amplitude are traveling in the same direction along a string. If they have a phase difference of π\piπ, what is the resulting amplitude of the superposed wave?

Correct Answer: A

Q23.Two waves with the same amplitude and frequency travel in opposite directions along a string and interfere to produce a standing wave. If the distance between two consecutive nodes is 0.5 m, what is the wavelength of the original waves?

Correct Answer: B

Q24.A pipe of length LLL is open at one end and closed at the other. It resonates at a fundamental frequency fff. If the pipe is now open at both ends, what will be the new fundamental frequency?

Correct Answer: B

Q25.Which of the following statements is true about longitudinal waves?

Correct Answer: C

Q26.A string fixed at both ends is vibrating in its fundamental mode. If the length of the string is LLL and the speed of the wave on the string is vvv, what is the frequency of the fundamental mode?

Correct Answer: A

Q27.In the context of waves, what is the principle of superposition?

Correct Answer: B

Q28.A pipe closed at one end resonates with a frequency of 300 Hz in its fundamental mode. If the speed of sound in air is 340 m/s, what is the length of the pipe?