WavesClass 11 Physics Notes

INTRODUCTION

Imagine dropping a pebble into a still pond. You see ripples, or disturbances, spreading out in circles. If you place a cork on the water, you'll notice it bobs up and down but doesn't travel outward with the ripples. This tells us something important: it's the disturbance that is moving, not the water itself. This moving disturbance, which transfers energy without transferring matter, is called a wave.

Waves are all around us and are essential for communication. When we speak, we create sound waves in the air. Our ears detect these waves. Modern communication often involves converting one type of wave into another. For example, a sound wave can be turned into an electrical signal, which then generates an electromagnetic wave to be sent via satellite.

Types of Waves

There are three main categories of waves:

1. Mechanical Waves: These are the most familiar types, like waves on a string, water waves, and sound waves. They need a medium (a substance like a solid, liquid, or gas) to travel through. They work because the particles in the medium are connected by elastic forces, allowing a disturbance to pass from one particle to the next. Mechanical waves cannot travel through a vacuum.

2. Electromagnetic Waves: These waves, which include light, radio waves, and X-rays, are different because they do not require a medium. They can travel through the vacuum of space. This is how light from distant stars reaches us. In a vacuum, all electromagnetic waves travel at the same incredible speed, known as the speed of light, c, which is $c=299,792,458 \text{ ms}^{-1}$.

3. Matter Waves: In the realm of quantum mechanics, particles like electrons, protons, and even atoms can behave like waves. These are called matter waves. While they are a more abstract concept, they have practical applications, such as in electron microscopes.

In this chapter, we will focus on mechanical waves.

How Mechanical Waves Work

Mechanical waves are closely related to oscillations. A medium that can support a wave, like a stretched string or the air, is an elastic medium. This means that if you disturb a part of it, restoring forces will try to bring it back to its equilibrium position.

Similarly, a sound wave travels through air by creating regions of compression (where air molecules are packed together) and rarefaction (where they are spread apart). As molecules in a compressed region push on their neighbors, they create a new compression, leaving a rarefaction behind them. This chain reaction allows the sound to travel without any net flow of air.

TRANSVERSE AND LONGITUDINAL WAVES

Mechanical waves can be classified into two types based on how the particles of the medium move relative to the direction of the wave's travel.

Transverse Waves

In a transverse wave, the particles of the medium oscillate perpendicular (at a right angle) to the direction of wave propagation.

Because transverse waves involve particles sliding past each other, they create a shearing strain in the medium. Therefore, transverse waves can only travel through media that can resist shearing stress, which are primarily solids. They cannot travel through fluids (liquids and gases).

Longitudinal Waves

In a longitudinal wave, the particles of the medium oscillate parallel to (along the same direction as) the direction of wave propagation.

Longitudinal waves involve compressing and expanding the medium. Since solids, liquids, and gases can all be compressed, longitudinal waves can travel through all elastic media.

DISPLACEMENT RELATION IN A PROGRESSIVE WAVE

A wave that travels from one point in a medium to another is called a progressive wave or a travelling wave. To describe such a wave mathematically, we need a function that tells us the displacement of any particle at any position x and at any time t.

For a sinusoidal transverse wave travelling in the positive x-direction, this function is: $y(x, t)=a \sin (k x-\omega t+\phi)$

Let's break down the components of this equation:

• y(x, t) is the displacement of the particle at position x and time t from its equilibrium position.
• a is the amplitude, which is the maximum displacement of a particle from its equilibrium position. The displacement y can be positive or negative, but the amplitude a is always positive.
• k is the angular wave number. It relates to how the wave varies with position.
• ω (omega) is the angular frequency. It relates to how the wave varies with time.
• ($k x-\omega t+\phi$) is the phase of the wave. It determines the state of oscillation (displacement and velocity) of a particle at a specific position and time.
• φ (phi) is the initial phase angle or phase constant. It represents the phase of the wave at x=0 and t=0.

Wavelength and Angular Wave Number

The wavelength, denoted by λ (lambda), is the minimum spatial distance between two points on a wave that are in the same phase. A simpler way to think about it is the distance between two consecutive crests (peaks) or two consecutive troughs (valleys).

The angular wave number k is related to the wavelength by the formula: $k=\frac{2 \pi}{\lambda}$ The SI unit for k is radians per metre ($\text{rad m}^{-1}$).

Period, Angular Frequency and Frequency

The period (T) is the time it takes for one particle in the medium to complete one full oscillation.

The angular frequency (ω) is related to the period by: $\omega=\frac{2 \pi}{T}$ Its SI unit is radians per second ($\text{rad s}^{-1}$).

The frequency (ν) is the number of complete oscillations a particle makes per second. It is the reciprocal of the period and is measured in hertz (Hz). $\nu=\frac{1}{T}=\frac{\omega}{2 \pi}$

For a longitudinal wave, we use the same concepts, but the displacement is often denoted by s(x, t) instead of y(x, t), as the displacement is along the x-axis: $s(x, t)=a \sin (k x-\omega t+\phi)$

Given

• Wave equation: $y(x, t)=0.005 \sin (80.0 x-3.0 t)$
• Standard form: $y(x, t)=a \sin (k x-\omega t)$
• Position, $x = 30.0 \text{ cm} = 0.3 \text{ m}$
• Time, $t = 20 \text{ s}$

To Find

(a) Amplitude, $a$ (b) Wavelength, $\lambda$ (c) Period, $T$, and frequency, $\nu$ (d) Displacement, $y$, at the given $x$ and $t$

Formula

$\lambda = \frac{2\pi}{k}$ $T = \frac{2\pi}{\omega}$ $\nu = \frac{1}{T}$

Solution

By comparing the given equation with the standard form, we can identify the parameters:

• Amplitude, $a = 0.005 \text{ m}$
• Angular wave number, $k = 80.0 \text{ rad m}^{-1}$
• Angular frequency, $\omega = 3.0 \text{ rad s}^{-1}$

(a) Amplitude

The amplitude of the wave is $a = 0.005 \text{ m} = 5 \text{ mm}$.

(b) Wavelength

We use the relationship between wavelength and angular wave number: $\lambda = \frac{2\pi}{k} = \frac{2\pi}{80.0 \text{ m}^{-1}} \approx 0.0785 \text{ m} = 7.85 \text{ cm}$

(c) Period and Frequency

First, we find the period from the angular frequency: $T = \frac{2\pi}{\omega} = \frac{2\pi}{3.0 \text{ s}^{-1}} \approx 2.09 \text{ s}$ Next, we find the frequency, which is the inverse of the period: $\nu = \frac{1}{T} = \frac{1}{2.09 \text{ s}} \approx 0.48 \text{ Hz}$

(d) Displacement at x = 30.0 cm and t = 20 s

We substitute the values of $x$ and $t$ into the wave equation: $y = (0.005 \text{ m}) \sin (80.0 \times 0.3 - 3.0 \times 20)$ $y = (0.005 \text{ m}) \sin (24 - 60) = (0.005 \text{ m}) \sin (-36)$ Here, the argument of the sine function is in radians. $y = (0.005 \text{ m}) \sin (-36 \text{ rad})$ To evaluate this, we can use a calculator. Note that $-36 \text{ rad} \approx 1.699 \text{ rad}$ within the principal range after removing multiples of $2\pi$. $y \approx (0.005 \text{ m}) \sin(1.699) \approx (0.005 \text{ m}) \times 0.9936 \approx 0.00497 \text{ m} \approx 5 \text{ mm}$

Final Answer (a) The amplitude is $5 \text{ mm}$. (b) The wavelength is $7.85 \text{ cm}$. (c) The period is $2.09 \text{ s}$ and the frequency is $0.48 \text{ Hz}$. (d) The displacement at the given point and time is approximately $5 \text{ mm}$.

THE SPEED OF A TRAVELLING WAVE

The speed of a wave is the speed at which the disturbance propagates through the medium. We can determine this by tracking a point of constant phase, like a crest.

The phase of a wave is given by the term $(k x-\omega t)$. For a point of constant phase: $k x-\omega t=\mathrm{constant}$ If we differentiate this expression with respect to time t, we get: $k \frac{dx}{dt} - \omega = 0$ The term $\frac{dx}{dt}$ is the speed of the wave, v. So, $v = \frac{dx}{dt}=\frac{\omega}{k}$

We can also express the wave speed in terms of wavelength and frequency. Substituting $\omega = 2\pi\nu$ and $k = 2\pi/\lambda$: $v=\frac{2 \pi \nu}{2 \pi / \lambda}=\lambda \nu$ This fundamental relationship, v = λν, holds true for all progressive waves. It means the speed of a wave is the product of its wavelength and frequency.

Speed of a Transverse Wave on Stretched String

For a transverse wave on a stretched string, the speed depends on two properties of the medium (the string):

1. Tension (T): This is the restoring force that pulls the string back to its equilibrium position. A higher tension results in a faster wave.
2. Linear Mass Density (μ): This is the mass per unit length of the string (μ = m/L). It represents the string's inertia. A heavier (denser) string resists motion more, resulting in a slower wave.

The formula for the speed of a transverse wave on a string is: $v=\sqrt{\frac{T}{\mu}}$

Given

• Length of wire, $L = 0.72 \text{ m}$
• Mass of wire, $m = 5.0 \times 10^{-3} \text{ kg}$
• Tension, $T = 60 \text{ N}$

To Find

Speed of the wave, $v$

Formula

First, find the linear mass density: $\mu = \frac{m}{L}$ Then, use the speed formula: $v=\sqrt{\frac{T}{\mu}}$

Solution

1. Calculate the linear mass density: $\mu = \frac{5.0 \times 10^{-3} \text{ kg}}{0.72 \text{ m}} = 6.9 \times 10^{-3} \text{ kg m}^{-1}$

2. Calculate the speed of the wave: $v=\sqrt{\frac{60 \text{ N}}{6.9 \times 10^{-3} \text{ kg m}^{-1}}} \approx \sqrt{8695.65} \approx 93 \text{ m s}^{-1}$

Final Answer The speed of transverse waves on the wire is approximately $93 \text{ m s}^{-1}$.

Speed of a Longitudinal Wave (Speed of Sound)

For a longitudinal wave, like sound, the speed depends on the medium's elastic properties and its density.

1. Elastic Modulus: This measures the medium's resistance to being deformed. For fluids (liquids and gases), this is the Bulk Modulus (B). For solids, it is often the Young's Modulus (Y). A higher modulus means the medium is "stiffer" and waves travel faster.
2. Density (ρ): This is the mass per unit volume. It represents the medium's inertia. Higher density leads to slower waves.

The general formula for the speed of a longitudinal wave is: $v=\sqrt{\frac{B}{\rho}} \quad \text{(for fluids)}$ $v=\sqrt{\frac{Y}{\rho}} \quad \text{(for a solid bar)}$

Generally, sound travels fastest in solids, then liquids, and slowest in gases. This is because solids and liquids are much harder to compress (have a much higher elastic modulus) than gases, and this effect outweighs their higher density.

Speed of Sound in a Gas: Newton's Formula and Laplace's Correction

Isaac Newton first derived a formula for the speed of sound in a gas. He assumed that the compressions and rarefactions caused by a sound wave happen so slowly that the temperature of the gas remains constant (an isothermal process). Under this assumption, the bulk modulus B is equal to the pressure P. This gives Newton's formula: $v=\sqrt{\frac{P}{\rho}}$

However, this formula gives a value for the speed of sound in air at Standard Temperature and Pressure (STP) of about $280 \text{ m s}^{-1}$, which is about 15% lower than the measured value of $331 \text{ m s}^{-1}$.

Pierre-Simon Laplace corrected this by pointing out that the compressions and rarefactions happen too quickly for heat to be exchanged with the surroundings. This means the process is adiabatic, not isothermal. For an adiabatic process, the bulk modulus is $B = \gamma P$, where γ (gamma) is the ratio of specific heats ($C_p/C_v$).

This leads to the Laplace correction, which is the correct formula for the speed of sound in a gas: $v=\sqrt{\frac{\gamma P}{\rho}}$ For air, $\gamma \approx 1.4$ (or 7/5), and using this value gives a calculated speed of sound that agrees very well with experimental results.

Given

• 1 mole of any gas occupies 22.4 litres ($22.4 \times 10^{-3} \text{ m}^3$) at STP.
• Mass of 1 mole of air, $M = 29.0 \times 10^{-3} \text{ kg}$.
• Standard pressure, $P = 1.01 \times 10^{5} \text{ N m}^{-2}$.
• Using Newton's formula.

To Find

The speed of sound in air at STP, $v$.

Formula

First, find the density of air at STP: $\rho_{o} = \frac{\text{mass of one mole}}{\text{volume of one mole at STP}}$ Then, use Newton's formula: $v=\sqrt{\frac{P}{\rho_{o}}}$

Solution

1. Calculate the density of air: $\rho_{o} = \frac{29.0 \times 10^{-3} \text{ kg}}{22.4 \times 10^{-3} \text{ m}^{3}} = 1.29 \text{ kg m}^{-3}$

2. Calculate the speed of sound using Newton's formula: $v=\sqrt{\frac{1.01 \times 10^{5} \text{ N m}^{-2}}{1.29 \text{ kg m}^{-3}}} \approx \sqrt{78294} \approx 280 \text{ m s}^{-1}$

Final Answer Using Newton's formula, the estimated speed of sound in air at STP is $280 \text{ m s}^{-1}$. (As noted above, this is incorrect, and the Laplace correction is needed for an accurate value.)

THE PRINCIPLE OF SUPERPOSITION OF WAVES

What happens when two or more waves travel through the same medium at the same time? The principle of superposition states that the net displacement of any particle in the medium is the algebraic sum of the displacements that would have been caused by each wave individually.

Mathematically, if two waves produce displacements $y_1(x, t)$ and $y_2(x, t)$, the resultant displacement $y(x, t)$ is: $y(x, t)=y_{1}(x, t)+y_{2}(x, t)$

Essentially, the waves pass through each other without being permanently altered. While they overlap, they combine to create a new wave shape.

Interference of Waves

When two harmonic waves with the same frequency and wavelength overlap, they create a phenomenon called interference.

Consider two waves travelling in the same direction with equal amplitude a but with a phase difference of φ: $y_{1}(x, t)=a \sin (k x-\omega t)$ $y_{2}(x, t)=a \sin (k x-\omega t+\phi)$

According to the principle of superposition, the resultant wave is: $y(x, t) = y_1 + y_2 = 2 a \cos \frac{\phi}{2} \sin \left(k x-\omega t+\frac{\phi}{2}\right)$

The resultant wave is also a harmonic wave with the same frequency and wavelength. However, its amplitude, $A = 2 a \cos(\phi/2)$, depends on the phase difference φ.

• Constructive Interference: When the waves are in phase ($\phi = 0, 2\pi, 4\pi, ...$), the cosine term is 1. The resultant amplitude is $A = 2a$, the maximum possible. The crests of one wave align with the crests of the other, reinforcing each other.
• Destructive Interference: When the waves are out of phase ($\phi = \pi, 3\pi, 5\pi, ...$), the cosine term is 0. The resultant amplitude is $A = 0$. The crests of one wave align with the troughs of the other, completely canceling each other out.

REFLECTION OF WAVES

When a wave encounters a boundary, it can be reflected. The nature of the reflection depends on the type of boundary.

Reflection at a Rigid Boundary

A rigid boundary is one that cannot move, like a string tied firmly to a wall. When a wave pulse reaches a rigid boundary, it gets reflected. The reflected wave has the same shape as the incident (incoming) wave but is inverted.

This inversion corresponds to a phase change of π radians (or 180°).

Why does this happen? At the rigid wall, the displacement must always be zero. By the principle of superposition, the only way for the incident pulse and reflected pulse to add up to zero at the wall is if the reflected pulse is an inverted version of the incident one.

If the incident wave is $y_{i}(x, t)=a \sin (k x-\omega t)$, the reflected wave will be: $y_{r}(x, t) = a \sin (k x+\omega t+\pi) = -a \sin (k x+\omega t)$

Reflection at an Open Boundary

An open boundary (or free end) is one that is free to move, like a string tied to a light ring that can slide frictionlessly up and down a rod.

When a wave pulse reaches an open boundary, it is reflected without any inversion. There is no phase change upon reflection.

The reflected wave has the same phase as the incident wave. If the incident wave is $y_{i}(x, t)=a \sin (k x-\omega t)$, the reflected wave will be: $y_{r}(x, t) = a \sin (k x+\omega t)$

STANDING WAVES AND NORMAL MODES

When a wave in a bounded medium (like a string fixed at both ends) reflects off a boundary, the reflected wave travels back and interferes with the original incident waves. The superposition of two identical waves travelling in opposite directions creates a special pattern called a standing wave or stationary wave.

Consider two waves with the same amplitude, frequency, and wavelength travelling in opposite directions: $y_{1}(x, t)=a \sin (k x-\omega t)$ $y_{2}(x, t)=a \sin (k x+\omega t)$

The resultant wave is: $y(x, t) = y_1 + y_2 = [2 a \sin k x] \cos \omega t$

Unlike a travelling wave, this wave pattern does not move left or right. Each particle simply oscillates up and down with an amplitude that depends on its position x.

Nodes and Antinodes

In a standing wave, there are specific points with distinct characteristics:

• Nodes: These are points where the amplitude is always zero ($2a \sin kx = 0$). The particles at these locations do not move at all. Nodes occur where $\sin kx = 0$, which means $kx = n\pi$ for $n = 0, 1, 2, ...$. The distance between two successive nodes is $\lambda/2$.
• Antinodes: These are points where the amplitude is maximum ($2a$). The particles at these locations oscillate with the largest possible amplitude. Antinodes occur where $|\sin kx| = 1$, which means $kx = (n+1/2)\pi$ for $n = 0, 1, 2, ...$. The distance between two successive antinodes is also $\lambda/2$.

Normal Modes of a Stretched String

For a string of length L fixed at both ends, the ends must be nodes. This boundary condition restricts the possible wavelengths and frequencies that can form standing waves on the string.

The condition for the end at $x=L$ to be a node is: $L = n \frac{\lambda}{2} \quad \text{for } n=1, 2, 3, ...$

This means only specific wavelengths are allowed: $\lambda_n=\frac{2 L}{n}$

The corresponding allowed frequencies, called natural frequencies or normal modes, are given by $v_n = v/\lambda_n$: $v_n=\frac{n v}{2 L}, \quad \text{for } n=1, 2, 3, ...$

• n = 1: Fundamental Frequency or First Harmonic. This is the lowest possible frequency of vibration. $v_1 = v/2L$.
• n = 2: Second Harmonic (or first overtone). $v_2 = 2(v/2L) = 2v_1$.
• n = 3: Third Harmonic (or second overtone). $v_3 = 3(v/2L) = 3v_1$.

A vibrating string, like on a guitar or violin, typically vibrates as a superposition of many of these harmonics, which gives the instrument its unique sound or timbre.

Normal Modes of an Air Column

Standing waves can also be set up in air columns, like in organ pipes or flutes.

Pipe Closed at One End

For a pipe of length L that is closed at one end and open at the other:

• The closed end must be a displacement node.
• The open end is a displacement antinode.

This boundary condition requires that the length of the pipe be an odd multiple of a quarter-wavelength: $L = (n+\frac{1}{2}) \frac{\lambda}{2} \quad \text{for } n=0, 1, 2, ...$ The allowed frequencies are: $v_n = (n+\frac{1}{2}) \frac{v}{2 L} = (2n+1) \frac{v}{4 L}$

• n = 0: Fundamental Frequency (First Harmonic): $v_1 = v/4L$.
• n = 1: Third Harmonic: $v_3 = 3(v/4L) = 3v_1$.
• n = 2: Fifth Harmonic: $v_5 = 5(v/4L) = 5v_1$.

Pipe Open at Both Ends

For a pipe of length L that is open at both ends, both ends must be displacement antinodes. This leads to the same frequency formula as a string fixed at both ends: $v_n=\frac{n v}{2 L}, \quad \text{for } n=1, 2, 3, ...$ An open pipe can produce all harmonics (both even and odd).

Given

• Length of pipe, $L = 30.0 \text{ cm} = 0.3 \text{ m}$
• Speed of sound, $v = 330 \text{ m s}^{-1}$
• Source frequency, $f_{source} = 1.1 \text{ kHz} = 1100 \text{ Hz}$

To Find

(i) Which harmonic mode resonates when the pipe is open at both ends. (ii) Whether resonance will occur if one end is closed.

Formula

For an open pipe: $v_n = \frac{nv}{2L}, \quad \text{for } n=1, 2, 3, ...$ For a closed pipe: $v_n = (2n+1)\frac{v}{4L}, \quad \text{for } n=0, 1, 2, ...$

Solution

(i) Pipe open at both ends

The possible harmonic frequencies for the open pipe are: $v_n = \frac{n \times 330 \text{ m s}^{-1}}{2 \times 0.3 \text{ m}} = \frac{n \times 330}{0.6} = 550n \text{ Hz}$ We need to find an integer n for which $v_n = 1100 \text{ Hz}$. $1100 = 550n$ $n = \frac{1100}{550} = 2$ Since n=2 is an integer, resonance occurs. This corresponds to the second harmonic.

(ii) Pipe closed at one end

The possible harmonic frequencies for the closed pipe are odd harmonics. The fundamental frequency is: $v_1 = \frac{v}{4L} = \frac{330 \text{ m s}^{-1}}{4 \times 0.3 \text{ m}} = \frac{330}{1.2} = 275 \text{ Hz}$ The other possible frequencies are odd multiples of this:

• $v_3 = 3 \times 275 = 825 \text{ Hz}$
• $v_5 = 5 \times 275 = 1375 \text{ Hz}$ The source frequency of $1100 \text{ Hz}$ is not one of the natural frequencies of the closed pipe ($1100 / 275 = 4$, which is an even multiple). Therefore, no resonance will be observed if one end of the pipe is closed.

BEATS

When two sound waves of slightly different frequencies are superposed, the intensity of the resulting sound is not constant. Instead, it varies periodically, creating a "waxing and waning" effect known as beats.

This phenomenon arises from the continuous cycling between constructive and destructive interference.

Consider two sound waves of equal amplitude a and slightly different angular frequencies $\omega_1$ and $\omega_2$: $s_{1}=a \cos \omega_{1} t$ $s_{2}=a \cos \omega_{2} t$

The resultant wave is: $s = s_1 + s_2 = \left[2 a \cos \omega_{b} t\right] \cos \omega_{a} t$ where:

• $\omega_{a}=\frac{(\omega_{1}+\omega_{2})}{2}$ is the average angular frequency.
• $\omega_{b}=\frac{(\omega_{1}-\omega_{2})}{2}$ is related to the beat frequency.

The resultant sound wave oscillates with the average frequency $\omega_a$, but its amplitude, $[2 a \cos \omega_{b} t]$, varies slowly over time. The intensity is maximum when $\cos \omega_b t = \pm 1$. This happens twice in each cycle of the cosine function. Therefore, the frequency at which the loudness waxes and wanes (the beat frequency) is $2\omega_b = \omega_1 - \omega_2$.

In terms of frequency ν (where $\omega = 2\pi\nu$), the beat frequency is simply the difference between the two source frequencies: $v_{\text{beat}} = |v_{1}-v_{2}|$

Musicians use beats to tune their instruments. By listening for the beat frequency between their instrument and a reference note, they can adjust the tension until the beats disappear ($v_{beat} = 0$), indicating the frequencies are identical.

Given

• Frequency of string A, $v_A = 427 \text{ Hz}$
• Initial beat frequency, $v_{beat,1} = 5 \text{ Hz}$
• Tension in string B is increased.
• Final beat frequency, $v_{beat,2} = 3 \text{ Hz}$

To Find

The original frequency of string B, $v_B$.

Formula

$v_{beat} = |v_A - v_B|$ Frequency of a string is related to tension by $v \propto \sqrt{T}$.

Solution

Initially, the beat frequency is 5 Hz. This means the frequency of B is either:

• Case 1: $v_B = v_A + 5 = 427 + 5 = 432 \text{ Hz}$
• Case 2: $v_B = v_A - 5 = 427 - 5 = 422 \text{ Hz}$

Now, the tension in string B is increased. Increasing the tension increases the frequency of the string ($v \propto \sqrt{T}$). Let the new frequency of B be $v_B'$. So, $v_B' > v_B$.

Let's analyze both cases:

• If Case 1 were true (original $v_B = 432 \text{ Hz}$): Increasing the tension would make $v_B'$ even higher (e.g., 433 Hz). The new beat frequency would be $|427 - v_B'|$, which would be greater than 5 Hz. This contradicts the observation that the beat frequency decreased to 3 Hz.

• If Case 2 were true (original $v_B = 422 \text{ Hz}$): Increasing the tension would make $v_B'$ higher than 422 Hz (e.g., 424 Hz). The new beat frequency would be $|427 - v_B'|$. As $v_B'$ increases from 422 Hz towards 427 Hz, the difference $|427 - v_B'|$ decreases. For example, if the new frequency is 424 Hz, the new beat frequency is $|427-424| = 3 \text{ Hz}$. This matches the observation.

Therefore, the original frequency of B must have been less than the frequency of A.

Final Answer The original frequency of B is $422 \text{ Hz}$.

Introduction to Waves

A wave is a disturbance that travels through a medium or space, transferring energy from one point to another without the physical transfer of matter. When you drop a pebble into a still pond, you see circular ripples spread out. A cork placed on the water will bob up and down but won't travel outward with the ripples. This shows that it's the disturbance, not the water itself, that is moving.

Waves are fundamental to our world. They are how we communicate, see, and hear.

• Speech and Hearing: Sound waves travel through the air from a speaker to a listener.
• Communication: Signals are transmitted through various types of waves, such as sound waves, electrical signals, and electromagnetic waves (like radio waves used by satellites).

There are three main categories of waves:

1. Mechanical Waves: These waves, like sound waves, water waves, and waves on a string, require a material medium (solid, liquid, or gas) to travel. They rely on the elastic properties of the medium, where particles oscillate and transfer energy to their neighbors. This chapter focuses on mechanical waves.
2. Electromagnetic Waves: These waves, including light, radio waves, and X-rays, do not require a medium and can travel through a vacuum. In a vacuum, all electromagnetic waves travel at the same speed, the speed of light, $c = 299,792,458 \text{ m s}^{-1}$.
3. Matter Waves: In quantum mechanics, particles like electrons, protons, and atoms are associated with waves. These matter waves are essential to technologies like the electron microscope.

The study of waves is built on the principles of oscillations. In an elastic medium like a stretched string or air, particles are bound by forces. When one particle is disturbed, it oscillates and passes that disturbance along to the next particle, creating a propagating wave.

Transverse and Longitudinal Waves

Mechanical waves are classified based on the direction of particle oscillation relative to the direction of wave propagation.

Transverse Waves

In a transverse wave, the particles of the medium oscillate perpendicular (at a right angle) to the direction the wave is travelling.

• Key Feature: Transverse waves involve a shearing strain, where layers of the medium slide past each other. Because of this, they can only travel through media that can resist shearing forces, such as solids. They cannot propagate through fluids (liquids and gases).

Longitudinal Waves

In a longitudinal wave, the particles of the medium oscillate back and forth along the same direction that the wave is travelling.

• Key Feature: Longitudinal waves involve compressive strain, changing the volume and density of the medium. Since solids, liquids, and gases can all be compressed, longitudinal waves can travel through all elastic media.

Combination Waves

Some waves are a mix of both transverse and longitudinal motion. For instance, the particles in an ocean wave move not just up and down but also back and forth, tracing a circular or elliptical path.

Answer

(a) Transverse and longitudinal: The sideways displacement creates a transverse wave, but the compression/stretching nature of the spring also allows for longitudinal motion. (b) Longitudinal: Pushing a piston back and forth in a liquid creates compressions and rarefactions, which is a longitudinal wave. (c) Transverse and longitudinal: Water waves created by a boat are a combination of both types of motion. (d) Longitudinal: Ultrasonic waves are sound waves (though at a frequency humans can't hear), and all sound waves in air are longitudinal.

Displacement Relation in a Progressive Wave

A progressive wave (or travelling wave) is a wave that moves from one point to another. To describe it mathematically, we need an equation that shows the displacement of any particle in the medium at any given time.

For a sinusoidal wave travelling along the x-axis, the displacement y at position x and time t is given by the equation: $y(x, t) = a \sin(kx - \omega t + \phi)$

Let's break down the components of this equation:

• $y(x, t)$: The displacement of the particle from its equilibrium position.
• $a$: The amplitude, which is the maximum displacement of a particle. The displacement y can be positive or negative, but the amplitude a is always positive.
• $(kx - \omega t + \phi)$: The phase of the wave. The phase determines the state of oscillation (displacement and velocity) of a particle at a specific position and time.
• $\phi$: The initial phase angle (or phase constant). It represents the phase of the wave at the origin ($x=0$) at the start time ($t=0$).
• $k$: The angular wave number (or propagation constant).
• $\omega$: The angular frequency.

• $(kx - \omega t)$ represents a wave travelling in the positive x-direction.
• $(kx + \omega t)$ represents a wave travelling in the negative x-direction.

Wavelength and Angular Wave Number

The wavelength ($\lambda$) is the minimum distance between two points on a wave that are in the same phase. For example, it's the distance from one crest to the next, or one trough to the next.

The angular wave number, $k$, is related to the wavelength by the formula: $k = \frac{2\pi}{\lambda}$ The SI unit for $k$ is radians per metre ($\text{rad m}^{-1}$).

Period, Angular Frequency, and Frequency

• The period (T) is the time it takes for one particle in the medium to complete one full oscillation.
• The angular frequency ($\omega$) is related to the period by: $\omega = \frac{2\pi}{T}$ Its SI unit is radians per second ($\text{rad s}^{-1}$).
• The frequency ($\nu$) is the number of complete oscillations per second. It is the reciprocal of the period and is related to the angular frequency by: $\nu = \frac{1}{T} = \frac{\omega}{2\pi}$ Frequency is measured in hertz (Hz).

For a longitudinal wave, the same principles apply, but the displacement is typically denoted by s instead of y: $s(x, t) = a \sin(kx - \omega t + \phi)$

Given

• Wave equation: $y(x, t) = 0.005 \sin(80.0x - 3.0t)$
• Position, $x = 30.0 \text{ cm} = 0.3 \text{ m}$
• Time, $t = 20 \text{ s}$

To Find

(a) Amplitude, $a$ (b) Wavelength, $\lambda$ (c) Period, $T$, and frequency, $\nu$ (d) Displacement, $y$, at the given $x$ and $t$

Formula

Standard wave equation: $y(x, t) = a \sin(kx - \omega t)$ Wavelength: $\lambda = 2\pi / k$ Period: $T = 2\pi / \omega$ Frequency: $\nu = 1/T$

Solution

By comparing the given equation with the standard form, we can identify the parameters:

• $a = 0.005 \text{ m}$
• $k = 80.0 \text{ rad m}^{-1}$
• $\omega = 3.0 \text{ rad s}^{-1}$

(a) Amplitude The amplitude of the wave is $a = 0.005 \text{ m}$, which is equal to $5 \text{ mm}$.

(b) Wavelength Using the relation between wavelength and angular wave number: $\lambda = \frac{2\pi}{k} = \frac{2\pi}{80.0 \text{ m}^{-1}} \approx 0.0785 \text{ m} = 7.85 \text{ cm}$

(c) Period and Frequency First, calculate the period using the angular frequency: $T = \frac{2\pi}{\omega} = \frac{2\pi}{3.0 \text{ s}^{-1}} \approx 2.09 \text{ s}$ Next, calculate the frequency: $\nu = \frac{1}{T} = \frac{1}{2.09 \text{ s}} \approx 0.48 \text{ Hz}$

(d) Displacement at $x = 0.3 \text{ m}$ and $t = 20 \text{ s}$ Substitute the values of $x$ and $t$ into the wave equation: $y = (0.005 \text{ m}) \sin(80.0 \times 0.3 - 3.0 \times 20)$ $y = (0.005 \text{ m}) \sin(24 - 60)$ $y = (0.005 \text{ m}) \sin(-36)$ The argument of the sine function is in radians. We can simplify this by noting that $\sin(\theta) = \sin(\theta + 2n\pi)$. Let's find an equivalent angle in a more familiar range. $12\pi \approx 37.699$. $\sin(-36) = \sin(-36 + 12\pi) \approx \sin(1.699 \text{ rad})$ Converting radians to degrees: $1.699 \text{ rad} \times \frac{180^\circ}{\pi} \approx 97^\circ$. $y \approx (0.005 \text{ m}) \sin(97^\circ) \approx 0.005 \text{ m} \times 0.9925 \approx 0.00496 \text{ m}$

Final Answer (a) Amplitude is $5 \text{ mm}$. (b) Wavelength is $7.85 \text{ cm}$. (c) Period is $2.09 \text{ s}$ and frequency is $0.48 \text{ Hz}$. (d) The displacement is approximately $5 \text{ mm}$.

The Speed of a Travelling Wave

The speed of a wave is the speed at which the disturbance propagates through the medium. We can determine this by tracking a point of constant phase, such as a crest.

For a point of constant phase, the term $(kx - \omega t)$ must be constant. $kx - \omega t = \text{constant}$ If we differentiate this with respect to time, we get: $k \frac{dx}{dt} - \omega = 0$ Since $\frac{dx}{dt}$ is the speed of the wave, $v$, we have: $v = \frac{dx}{dt} = \frac{\omega}{k}$

We can express this in terms of wavelength and frequency. Since $\omega = 2\pi\nu$ and $k = 2\pi/\lambda$: $v = \frac{2\pi\nu}{2\pi/\lambda} = \lambda\nu$ This gives us the fundamental wave speed relation: $v = \lambda \nu = \frac{\lambda}{T}$

This equation shows that in one period ($T$), the wave travels a distance equal to one wavelength ($\lambda$).

Speed of a Transverse Wave on a Stretched String

For a transverse wave on a string, the speed depends on two factors:

1. Tension ($T$): The restoring force that pulls the string back to its straight form. Higher tension means a faster wave.
2. Linear Mass Density ($\mu$): The mass per unit length of the string ($m/L$). This is the inertial property. A heavier string (higher $\mu$) means a slower wave.

The formula for the speed of a transverse wave on a string is: $v = \sqrt{\frac{T}{\mu}}$

Given

• Length, $L = 0.72 \text{ m}$
• Mass, $m = 5.0 \times 10^{-3} \text{ kg}$
• Tension, $T = 60 \text{ N}$

To Find

Speed of the wave, $v$

Formula

$v = \sqrt{\frac{T}{\mu}}$ where $\mu = m/L$

Solution

First, calculate the linear mass density, $\mu$: $\mu = \frac{5.0 \times 10^{-3} \text{ kg}}{0.72 \text{ m}} \approx 6.9 \times 10^{-3} \text{ kg m}^{-1}$

Now, substitute the values of $T$ and $\mu$ into the speed formula: $v = \sqrt{\frac{60 \text{ N}}{6.9 \times 10^{-3} \text{ kg m}^{-1}}} \approx \sqrt{8695.65} \approx 93 \text{ m s}^{-1}$

Final Answer The speed of transverse waves on the wire is approximately $93 \text{ m s}^{-1}$.

Speed of a Longitudinal Wave (Speed of Sound)

For a longitudinal wave, like sound, the speed depends on:

1. Bulk Modulus ($B$): The elastic property. It measures how resistant a substance is to compression. A higher bulk modulus (harder to compress) means a faster wave.
2. Density ($\rho$): The mass per unit volume. This is the inertial property. Higher density means a slower wave.

The general formula for the speed of a longitudinal wave in a fluid is: $v = \sqrt{\frac{B}{\rho}}$

For a solid bar, the relevant elastic property is the Young's Modulus ($Y$): $v = \sqrt{\frac{Y}{\rho}}$

Sound travels faster in solids and liquids than in gases because their bulk modulus values are much higher, which more than compensates for their greater densities.

Speed of Sound in a Gas: Newton's Formula and Laplace's Correction

Newton's Formula Isaac Newton first derived a formula for the speed of sound in a gas. He assumed that the compressions and rarefactions caused by a sound wave occur slowly enough for the process to be isothermal (constant temperature). For an ideal gas under isothermal conditions, the bulk modulus is equal to the pressure ($B = P$). This led to Newton's formula: $v = \sqrt{\frac{P}{\rho}}$ However, this formula gives a value for the speed of sound in air that is about 15% lower than the experimental value.

Laplace's Correction Pierre-Simon Laplace corrected Newton's assumption. He pointed out that the compressions and rarefactions happen so quickly that there is no time for heat to flow between them. The process is adiabatic (no heat exchange), not isothermal.

For an adiabatic process in an ideal gas, the bulk modulus is given by $B_{ad} = \gamma P$, where $\gamma$ (gamma) is the ratio of specific heats ($C_p/C_v$). This leads to the corrected formula, known as the Laplace formula: $v = \sqrt{\frac{\gamma P}{\rho}}$ For air, $\gamma \approx 1.4$ (or $7/5$), and this formula gives a value for the speed of sound that agrees very well with experimental results.

Given

• At STP, Pressure $P = 1.01 \times 10^5 \text{ N m}^{-2}$
• Mass of one mole of air = $29.0 \times 10^{-3} \text{ kg}$
• Volume of one mole of gas at STP = $22.4 \text{ litres} = 22.4 \times 10^{-3} \text{ m}^3$

To Find

Speed of sound in air, $v$, using Newton's formula.

Formula

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

Solution

First, calculate the density of air at STP: $\rho_o = \frac{\text{mass of one mole}}{\text{volume of one mole}} = \frac{29.0 \times 10^{-3} \text{ kg}}{22.4 \times 10^{-3} \text{ m}^3} \approx 1.29 \text{ kg m}^{-3}$

Now, use Newton's formula: $v = \sqrt{\frac{1.01 \times 10^5 \text{ N m}^{-2}}{1.29 \text{ kg m}^{-3}}} \approx \sqrt{78294} \approx 280 \text{ m s}^{-1}$

Final Answer Newton's formula estimates the speed of sound in air at STP to be $280 \text{ m s}^{-1}$. (Using Laplace's correction with $\gamma = 1.4$, the calculated speed is $v = \sqrt{1.4} \times 280 \approx 331.3 \text{ m s}^{-1}$, which is very close to the measured value of $331 \text{ m s}^{-1}$).

The Principle of Superposition of Waves

What happens when two or more waves meet at the same point in a medium? The Principle of Superposition states that the resultant displacement of a particle is the algebraic sum of the displacements that each individual wave would have caused. $y(x, t) = y_1(x, t) + y_2(x, t) + \dots$ After passing through each other, the waves continue on their way unchanged.

This principle leads to the phenomenon of interference. Let's consider two harmonic waves travelling in the same direction with the same amplitude ($a$), angular frequency ($\omega$), and wave number ($k$), but with a phase difference of $\phi$. $y_1(x, t) = a \sin(kx - \omega t)$ $y_2(x, t) = a \sin(kx - \omega t + \phi)$

Using the principle of superposition, the resultant displacement is: $y(x, t) = y_1 + y_2 = a[\sin(kx - \omega t) + \sin(kx - \omega t + \phi)]$ Using the trigonometric identity $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$, we get: $y(x, t) = \left(2a \cos \frac{\phi}{2}\right) \sin\left(kx - \omega t + \frac{\phi}{2}\right)$ This resultant wave has the same frequency and wavelength, but its amplitude, $A = 2a \cos(\phi/2)$, depends on the phase difference $\phi$.

• Constructive Interference: Occurs when the waves are in phase ($\phi = 0, 2\pi, 4\pi, \dots$). Here, $\cos(\phi/2) = 1$, and the resultant amplitude is maximum, $A = 2a$. The waves add up.
• Destructive Interference: Occurs when the waves are out of phase ($\phi = \pi, 3\pi, 5\pi, \dots$). Here, $\cos(\phi/2) = 0$, and the resultant amplitude is zero, $A = 0$. The waves cancel each other out.

Reflection of Waves

When a travelling wave encounters a boundary, it can be reflected. The nature of the reflection depends on the type of boundary.

Reflection at a Rigid Boundary

When a wave on a string hits a fixed end (like a wall), it reflects back. The reflected wave is inverted. This corresponds to a phase change of $\pi$ (or 180°).

• If the incident wave is $y_i(x, t) = a \sin(kx - \omega t)$,
• The reflected wave travelling in the opposite direction is $y_r(x, t) = a \sin(kx + \omega t + \pi) = -a \sin(kx + \omega t)$.

This happens because at a rigid boundary, the displacement must always be zero. By the principle of superposition, the incident and reflected waves must cancel each other out at the boundary at all times.

Reflection at an Open Boundary

When a wave on a string hits a free end (like a ring that can slide frictionlessly on a rod), it reflects back without being inverted. There is no phase change upon reflection.

• If the incident wave is $y_i(x, t) = a \sin(kx - \omega t)$,
• The reflected wave is $y_r(x, t) = a \sin(kx + \omega t)$.

Standing Waves and Normal Modes

When two identical waves travelling in opposite directions interfere, they can create a standing wave or stationary wave. This often happens when a wave is reflected from a boundary and interferes with the incoming wave.

Consider two waves:

• Incident wave: $y_1(x, t) = a \sin(kx - \omega t)$
• Reflected wave: $y_2(x, t) = a \sin(kx + \omega t)$

The superposition gives: $y(x, t) = y_1 + y_2 = a[\sin(kx - \omega t) + \sin(kx + \omega t)]$ Using a trigonometric identity, this simplifies to: $y(x, t) = [2a \sin(kx)] \cos(\omega t)$

This is the equation for a standing wave. Notice two key features:

1. The wave pattern does not travel; it oscillates in place.
2. The amplitude of oscillation, $[2a \sin(kx)]$, is not constant but depends on the position $x$.

Nodes and Antinodes

• Nodes are points on the standing wave that have zero amplitude and remain stationary at all times. They occur where $\sin(kx) = 0$, which means $kx = n\pi$ for $n = 0, 1, 2, \dots$. The distance between two consecutive nodes is $\lambda/2$.
• Antinodes are points that have the maximum possible amplitude ($2a$). They occur where $|\sin(kx)| = 1$, which means $kx = (n + 1/2)\pi$. The distance between two consecutive antinodes is also $\lambda/2$.

Normal Modes of a Stretched String

For a string of length $L$ fixed at both ends ($x=0$ and $x=L$), the ends must be nodes. This condition restricts the possible wavelengths and frequencies of the standing waves that can exist on the string. The condition for the far end to be a node is $L = n(\lambda/2)$, where $n=1, 2, 3, \dots$.

This means the allowed wavelengths are: $\lambda_n = \frac{2L}{n}$ The corresponding allowed frequencies, called the normal modes or harmonics, are: $\nu_n = \frac{v}{\lambda_n} = \frac{nv}{2L} \quad (n=1, 2, 3, \dots)$

• $n=1$: Fundamental frequency or first harmonic. This is the lowest possible frequency.
• $n=2$: Second harmonic (or first overtone).
• $n=3$: Third harmonic (or second overtone), and so on.

Normal Modes of an Air Column

Standing waves can also be set up in air columns, such as in organ pipes or flutes. The boundary conditions are different:

• A closed end of a pipe is a displacement node.
• An open end of a pipe is a displacement antinode.

1. Pipe Closed at One End, Open at the Other Let the length of the pipe be $L$. There is a node at the closed end ($x=0$) and an antinode at the open end ($x=L$). This requires that the length of the pipe be an odd multiple of a quarter-wavelength: $L = (n + 1/2)(\lambda/2)$ for $n=0, 1, 2, \dots$.

The allowed frequencies are: $\nu_n = (n + 1/2)\frac{v}{2L} = (2n+1)\frac{v}{4L} \quad (n=0, 1, 2, \dots)$

• The fundamental frequency ($n=0$) is $\nu_0 = v/4L$.
• Only odd harmonics ($v/4L, 3v/4L, 5v/4L, \dots$) are present.

2. Pipe Open at Both Ends There is an antinode at both ends. This requires the length of the pipe to be an integer multiple of a half-wavelength: $L = n(\lambda/2)$.

The allowed frequencies are: $\nu_n = \frac{nv}{2L} \quad (n=1, 2, 3, \dots)$

• All harmonics are present, just like on a string fixed at both ends.

Given

• Length of pipe, $L = 30.0 \text{ cm} = 0.3 \text{ m}$
• Source frequency, $\nu_{source} = 1.1 \text{ kHz} = 1100 \text{ Hz}$
• Speed of sound, $v = 330 \text{ m s}^{-1}$

To Find

(a) The harmonic mode n for an open pipe that resonates with the source. (b) Whether resonance will occur if one end is closed.

Formula

• For an open pipe: $\nu_n = \frac{nv}{2L}$ for $n=1, 2, 3, \dots$
• For a pipe closed at one end: $\nu_n = (2n+1)\frac{v}{4L}$ for $n=0, 1, 2, \dots$

Solution

(a) Open Pipe The frequencies of the normal modes for the open pipe are: $\nu_n = \frac{n(330 \text{ m s}^{-1})}{2(0.3 \text{ m})} = \frac{n \times 330}{0.6} = 550n \text{ Hz}$ We need to find an integer n for which $\nu_n = 1100 \text{ Hz}$. $1100 = 550n$ $n = \frac{1100}{550} = 2$ Resonance occurs at the second harmonic.

(b) Pipe Closed at One End Now, let's find the possible resonant frequencies if one end is closed. The fundamental frequency is: $\nu_0 = \frac{v}{4L} = \frac{330 \text{ m s}^{-1}}{4(0.3 \text{ m})} = \frac{330}{1.2} = 275 \text{ Hz}$ Only odd multiples of this fundamental frequency are possible modes:

• 1st harmonic ($n=0$): $1 \times 275 = 275 \text{ Hz}$
• 3rd harmonic ($n=1$): $3 \times 275 = 825 \text{ Hz}$
• 5th harmonic ($n=2$): $5 \times 275 = 1375 \text{ Hz}$ The source frequency of $1100 \text{ Hz}$ is not an odd multiple of the fundamental frequency ($1100/275 = 4$, which is an even number). Therefore, it does not correspond to a possible mode for the closed pipe.

Final Answer (a) The 1.1 kHz source will resonate with the second harmonic of the open pipe. (b) No resonance will be observed with the same source if one end of the pipe is closed.

Beats

Beats are the periodic and repeating fluctuations in the loudness of a sound that occur when two sound waves of very similar (but not identical) frequencies interfere with each other. The loudness of the sound waxes and wanes.

This phenomenon is a direct result of superposition. Consider two sound waves with equal amplitude a and slightly different angular frequencies $\omega_1$ and $\omega_2$. At a fixed point ($x=0$), their displacements are: $s_1 = a \cos(\omega_1 t)$ $s_2 = a \cos(\omega_2 t)$ The resultant displacement is: $s = s_1 + s_2 = a(\cos(\omega_1 t) + \cos(\omega_2 t))$ Using a trigonometric identity, this becomes: $s = \left[2a \cos\left(\frac{\omega_1 - \omega_2}{2}t\right)\right] \cos\left(\frac{\omega_1 + \omega_2}{2}t\right)$ This can be interpreted as a wave oscillating rapidly at the average frequency, $\omega_a = (\omega_1 + \omega_2)/2$, but with a slowly varying amplitude given by the term in the brackets.

The loudness is maximum when the amplitude term is maximum, which occurs twice in each cycle of the cosine function. The frequency of this amplitude variation (the beat) is twice the frequency of the amplitude term, $\omega_b = (\omega_1 - \omega_2)/2$. The angular frequency of the beats is $2\omega_b = \omega_1 - \omega_2$.

The beat frequency, $\nu_{beat}$, is the difference between the two original frequencies: $\nu_{beat} = |\nu_1 - \nu_2|$

Given

• Frequency of string A, $\nu_A = 427 \text{ Hz}$
• Initial beat frequency, $|\nu_A - \nu_B| = 5 \text{ Hz}$
• Final beat frequency (after increasing tension in B) = 3 Hz

To Find

The original frequency of string B, $\nu_B$.

Formula

• Beat frequency: $\nu_{beat} = |\nu_1 - \nu_2|$
• Wave speed on a string: $v = \sqrt{T/\mu}$
• Frequency of a string: $\nu \propto v \propto \sqrt{T}$

Solution

From the initial condition, we have two possibilities for the original frequency of B:

1. $\nu_A - \nu_B = 5 \implies \nu_B = 427 - 5 = 422 \text{ Hz}$
2. $\nu_B - \nu_A = 5 \implies \nu_B = 427 + 5 = 432 \text{ Hz}$

Now, consider the effect of increasing the tension in string B. Increasing the tension increases the wave speed on the string, which in turn increases its frequency ($\nu \propto \sqrt{T}$). So, $\nu_B$ increases.

Let's test both possibilities:

• Case 1: Original $\nu_B = 422 \text{ Hz}$. When tension is increased, $\nu_B$ increases, getting closer to $\nu_A$ (427 Hz). This would cause the difference $|\nu_A - \nu_B|$ (the beat frequency) to decrease. This matches the observation that the beat frequency decreased from 5 Hz to 3 Hz.
• Case 2: Original $\nu_B = 432 \text{ Hz}$. When tension is increased, $\nu_B$ increases further, moving away from $\nu_A$ (427 Hz). This would cause the difference $|\nu_A - \nu_B|$ to increase. This contradicts the observation.

Therefore, the first case must be correct.

Final Answer The original frequency of string B is $422 \text{ Hz}$.