Step 1: Analyzing the Kinematics of the Falling Stone

The motion of the stone is governed by the equations of motion under constant acceleration due to gravity. The time taken ($t_1$) for the stone to reach the water surface is calculated using the second equation of motion: $s = ut + \frac{1}{2}at^2$.

• Initial velocity ($u$) = $0 \text{ m/s}$ (since it is dropped)
• Displacement ($s$) = $300 \text{ m}$
• Acceleration ($g$) = $9.8 \text{ m/s}^2$

Substituting these values into the equation: $300 = 0 \cdot t_1 + \frac{1}{2} \cdot 9.8 \cdot t_1^2$ $300 = 4.9 \cdot t_1^2$ $t_1^2 = \frac{300}{4.9} \approx 61.224$ $t_1 = \sqrt{61.224} \approx 7.825 \text{ s}$

Step 2: Analyzing the Propagation of Sound

Once the stone hits the water, the sound travels back to the top of the tower. This is a motion at a constant velocity ($v_s = 340 \text{ m/s}$). The time taken ($t_2$) for the sound to travel the same distance is given by: $t_2 = \frac{s}{v_s}$

$t_2 = \frac{300}{340} \approx 0.882 \text{ s}$

Step 3: Calculating Total Time

The total time ($T$) elapsed between dropping the stone and hearing the splash is the sum of the time taken for the stone to fall and the time taken for the sound to return [Per the principle of superposition of time intervals]: $T = t_1 + t_2$ $T = 7.825 \text{ s} + 0.882 \text{ s}$ $T = 8.707 \text{ s}$

Final Solution: The splash is heard at the top of the tower approximately 8.71 seconds after the stone is dropped.

Step 1: Theoretical Foundation of Resonance in a Pipe

For a pipe open at one end and closed at the other (by a movable piston), resonance occurs at odd harmonics of the fundamental frequency. The condition for resonance is given by the length of the air column $L$ and the wavelength $\lambda$ as:

$L_n = (2n - 1) \frac{\lambda}{4}$, where $n = 1, 2, 3, \dots$

[Per the principle of standing waves in a tube with one open end, nodes occur at the closed end and antinodes at the open end.]

Step 2: Identifying Consecutive Resonance Positions

Given the two consecutive resonant lengths:

• $L_1 = 25.5 \text{ cm} = 0.255 \text{ m}$
• $L_2 = 79.3 \text{ cm} = 0.793 \text{ m}$

Step 3: Calculation of the Speed of Sound

Using the wave equation relation $v = f\lambda$, where $f$ is the frequency of the source and $v$ is the wave velocity:
Given frequency $f = 340 \text{ Hz}$.
$v = 340 \text{ s}^{-1} \times 1.076 \text{ m}$
$v = 365.84 \text{ m/s}$

Step 4: Visualizing the Standing Wave

Step 5: Analysis of Precision

The speed of sound $v$ is determined by the relationship between the wavelength derived from the physical nodes in the tube and the fixed frequency of the tuning fork.

Final Solution: The estimated speed of sound in air at the temperature of the experiment is 365.84 m/s.

Step 1: Analyzing the Initial Condition

The beat frequency $f_b$ is defined as the absolute difference between the frequencies of two sounding sources: $f_b = |f_A - f_B|$.

Given the initial beat frequency is $6\text{ Hz}$, we have:

$|f_A - f_B| = 6\text{ Hz}$

Given $f_A = 324\text{ Hz}$, the possible values for $f_B$ are:

• $f_B = 324 + 6 = 330\text{ Hz}$
• $f_B = 324 - 6 = 318\text{ Hz}$

Step 2: Understanding the Effect of Tension on Frequency

The fundamental frequency $f$ of a stretched string is given by the formula:

$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$

Where $T$ is tension, $L$ is length, and $\mu$ is linear mass density. [Per Mersenne's Laws of vibrating strings].

Since $f \propto \sqrt{T}$, reducing the tension $T$ in string $A$ leads to a decrease in the frequency $f_A$. Therefore, the new frequency $f_A'$ is less than $324\text{ Hz}$.

Step 3: Evaluating the Scenarios with Reduced Tension

We are given that after decreasing $T_A$, the beat frequency reduces to $3\text{ Hz}$. Let $f_A' < 324\text{ Hz}$.

In Case 1, as $f_A$ decreases from $324\text{ Hz}$ toward $330\text{ Hz}$, the difference $|324 - 330| = 6\text{ Hz}$ increases (e.g., if $f_A' = 320$, difference is $10$).

In Case 2, as $f_A$ decreases from $324\text{ Hz}$ toward $318\text{ Hz}$, the difference $|f_A - 318|$ decreases. If $f_A$ changes from $324$ to $321$, the beat frequency becomes $|321 - 318| = 3\text{ Hz}$. This matches the experimental observation.

Step 4: Logical Deduction

Since the beat frequency reduced from $6\text{ Hz}$ to $3\text{ Hz}$ upon decreasing the frequency of string A, the initial frequency of string A must have been higher than the frequency of string B.

Therefore, $f_A - f_B = 6\text{ Hz}$.

$324 - f_B = 6 \implies f_B = 318\text{ Hz}$.

Final Solution: The frequency of string B is 318 Hz.

Given Variables & Initial Setup

The wave on a string described in the previous context (Exercise 15.11) is defined by the traveling wave equation:

$y(x, t) = 0.06 \sin(2\pi x + 3\pi t + \pi/4)$

Where:

• $A = 0.06 \, \text{m}$ (Amplitude)
• $k = 2\pi \, \text{rad/m}$ (Wave number)
• $\omega = 3\pi \, \text{rad/s}$ (Angular frequency)

Step 1: Analyzing the Amplitude of a Traveling Wave

A traveling wave is mathematically represented by the general form $y(x, t) = A \sin(kx \pm \omega t + \phi)$. In this expression, $A$ represents the maximum displacement of any particle from its mean position [Per the Theory of Simple Harmonic Motion].

Step 2: Evaluating Spatial Dependence

To determine if the amplitude varies with position $x$, we examine the partial derivative of $y$ with respect to time $t$ to find the velocity, or more simply, analyze the functional form of the displacement $y(x, t)$.

In the equation $y(x, t) = 0.06 \sin(2\pi x + 3\pi t + \pi/4)$, the term $0.06$ is a constant coefficient [By the algebraic properties of sinusoidal functions]. Regardless of the values of $x$ (position) or $t$ (time), the peak value of the sine function $\sin(\theta)$ is always $1$.

Consequently, the maximum displacement (amplitude) for any point $x$ on the string is given by:

$y_{\text{max}} = |A| \cdot |\sin(kx + \omega t + \phi)|_{\text{max}} = 0.06 \times 1 = 0.06 \, \text{m}$

Step 3: Comparative Analysis (Visual Representation)

Step 4: Justification

Since the amplitude $A$ is independent of the spatial coordinate $x$ and the temporal coordinate $t$, every point on the string undergoes simple harmonic motion with an identical maximum displacement [Per the definition of a monochromatic plane wave].

Final Solution: Yes, all points on the string oscillate with the same amplitude of 0.06 meters, as the given equation describes a uniform traveling wave where the amplitude is independent of the position x.

Given Variables & Initial Setup

To determine the wavelength of the ultrasonic wave in the biological tissue, we identify the following physical parameters provided in the problem statement:

• Speed of sound ($v$) = $1.7\text{ km s}^{-1} = 1.7 \times 10^{3}\text{ m s}^{-1}$ [Conversion to SI units]
• Frequency of the scanner ($f$) = $4.2\text{ MHz} = 4.2 \times 10^{6}\text{ Hz}$ [Standard SI unit conversion]

Step 1: Establishing the Wave Equation

The fundamental relationship between the speed of a wave, its frequency, and its wavelength is governed by the wave equation [Per the linear wave propagation theory]:

$v = f \cdot \lambda$

Where:

• $v$ is the velocity of the wave in the medium.
• $f$ is the frequency of the oscillation.
• $\lambda$ is the wavelength.

Step 2: Algebraic Rearrangement

To isolate the target variable, wavelength ($\lambda$), we rearrange the equation as follows:

$\lambda = \frac{v}{f}$

Step 3: Numerical Substitution & Calculation

Substituting the given values into the derived formula:

$\lambda = \frac{1.7 \times 10^{3}\text{ m s}^{-1}}{4.2 \times 10^{6}\text{ s}^{-1}}$

$\lambda = \left( \frac{1.7}{4.2} \right) \times 10^{3-6}\text{ m}$

$\lambda \approx 0.40476 \times 10^{-3}\text{ m}$

Step 4: Unit Conversion & Final Precision

To express the result in more standard units for biological imaging (millimeters):

$\lambda \approx 4.0476 \times 10^{-4}\text{ m}$

$\lambda \approx 0.405 \times 10^{-3}\text{ m} = 0.405\text{ mm}$

Visual Representation of Wave Parameters

Final Solution: The wavelength of the ultrasonic sound in the tissue is approximately $4.05 \times 10^{-4}\text{ m}$ or $0.405\text{ mm}$.

Given Variables & Initial Setup

The system involves a steel rod undergoing longitudinal vibrations. The fundamental parameters are provided as follows:

• Length of the rod, $L = 100\text{ cm} = 1.0\text{ m}$
• Fundamental frequency, $\nu = 2.53\text{ kHz} = 2530\text{ Hz}$
• Boundary Condition: The rod is clamped at its middle (a node is formed at the center).

Step 1: Analyzing Waveform Dynamics

When a rod is clamped at its midpoint, the midpoint acts as a stationary node ($N$). Because the ends of the rod are free, they must act as antinodes ($A$). The simplest longitudinal vibration pattern (the fundamental mode) under these constraints corresponds to the structure $A—N—A$.

In this configuration, the distance between two consecutive antinodes is half a wavelength ($\frac{\lambda}{2}$), but here, the length of the rod $L$ represents the distance from one end to the center and the center to the other end. Specifically, the length $L$ corresponds to exactly half of the wavelength ($\lambda/2$) of the fundamental vibration, because the rod is essentially a pipe open at both ends in the context of standing longitudinal waves.

[Per the Standing Wave boundary condition for a rod clamped at the center]:

$L = \frac{\lambda}{2} \implies \lambda = 2L$

Step 2: Mathematical Derivation

We utilize the fundamental wave equation relating speed ($v$), frequency ($\nu$), and wavelength ($\lambda$):

$v = \nu \lambda$

Substituting the relation for wavelength derived in Step 1 ($\lambda = 2L$):

$v = \nu \times (2L)$

Step 3: Numerical Computation

Substitute the given values into the derived equation:

$v = 2530\text{ Hz} \times (2 \times 1.0\text{ m})$

$v = 2530 \times 2\text{ m/s}$

$v = 5060\text{ m/s}$

Visual Representation of Vibration Mode

Justification of Result

The calculation is based on the standard acoustic properties of longitudinal waves in a thin elastic rod, where the speed is independent of frequency and governed by the Young's modulus ($Y$) and density ($\rho$) of the material, defined as $v = \sqrt{Y/\rho}$. The result $5060\text{ m/s}$ is physically consistent with the speed of sound in common structural steel (typically ranging between $5000\text{--}5200\text{ m/s}$).

Final Solution: The speed of sound in the steel rod is $5060\text{ m/s}$.

Step 1: Given Variables & Initial Setup

To determine the apparent frequency of a sound wave when the source is in motion relative to a stationary observer, we apply the Doppler Effect formula. The variables provided are:

• Frequency of the source ($ν_0$): $400\text{ Hz}$
• Speed of sound in still air ($v$): $340\text{ m s}^{-1}$ [Standard atmospheric assumption for Physics problems]
• Velocity of the source ($v_s$): $10\text{ m s}^{-1}$
• Velocity of the observer ($v_o$): $0\text{ m s}^{-1}$ [The platform observer is stationary]

Step 2: Selecting the Doppler Effect Equation

The general formula for the frequency perceived by an observer ($ν'$) is given by:

$ν' = ν_0 \left( \frac{v + v_o}{v - v_s} \right)$

[Per the Doppler Effect, when the source approaches the observer, the effective wavelength decreases, causing an upward shift in observed frequency.]

Step 3: Execution of Calculation

Substitute the known values into the equation:

$ν' = 400 \left( \frac{340 + 0}{340 - 10} \right)$

$ν' = 400 \left( \frac{340}{330} \right)$

$ν' = 400 \left( 1.0303 \right)$

$ν' \approx 412.12\text{ Hz}$

Step 4: Visualizing the Wave Compression

Step 5: Analytical Summary

Final Solution: The frequency of the whistle heard by the platform observer is approximately 412.12 Hz.

Step 1: Given Variables & Theoretical Foundation

To determine the speed of sound in the steel rod, we identify the following parameters:

• Length of the steel rod ($L$): $100\text{ cm} = 1.0\text{ m}$
• Fundamental frequency ($f$): $2.53\text{ kHz} = 2.53 \times 10^3\text{ Hz}$
• Clamping condition: The rod is clamped at its middle.

[Per the theory of longitudinal waves in a rod clamped at the center, the center acts as a node ($N$) and the free ends act as antinodes ($A$).]

Step 2: Analysis of Standing Wave Configuration

When a rod is clamped at its midpoint, the fundamental mode of vibration requires a node at the center ($x = L/2$) and antinodes at both ends ($x = 0$ and $x = L$). The distance between two adjacent antinodes is $\lambda/2$. Here, the total length $L$ corresponds to two segments, each of length $\lambda/4$, or more simply, the rod constitutes a half-wavelength resonance.

The relationship between the length $L$ and the wavelength $\lambda$ for the fundamental frequency in this configuration is:

$L = \frac{\lambda}{2} \implies \lambda = 2L$ [By the superposition principle for longitudinal standing waves]

Step 3: Calculating the Speed of Sound

The fundamental frequency formula for longitudinal waves is defined as:

$v = f \cdot \lambda$

Substituting the geometric constraint $\lambda = 2L$ into the wave equation:

$v = f \cdot (2L)$

Substitute the given numerical values:

$v = (2.53 \times 10^3\text{ Hz}) \times (2 \times 1.0\text{ m})$

$v = 2.53 \times 10^3 \times 2$

$v = 5.06 \times 10^3\text{ m/s}$

$v = 5060\text{ m/s}$

Step 4: Visual Representation of the Mode

Step 5: Final Evaluation

Final Solution: The speed of sound in the steel rod is $5060\text{ m/s}$.

Given Variables & Initial Setup

Consider a standard traveling harmonic wave described by the function:

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

For this specific analysis, let us adopt the standard parameters: amplitude $a = 1 \text{ cm}$, angular frequency $\omega = 2\pi \text{ rad/s}$, and wave number $k = \pi/2 \text{ cm}^{-1}$ (assuming $\phi = 0$ for simplicity). The displacement function is therefore: $y(x, t) = \sin\left(\frac{\pi}{2}x - 2\pi t\right)$.

Step 1: Displacement vs. Time Equations for Specific Positions

To determine the displacement $y$ as a function of time $t$ for specific points $x$, we substitute the given values into the wave equation:

• For $x = 0$ cm: $y(0, t) = \sin(0 - 2\pi t) = -\sin(2\pi t)$
• For $x = 2$ cm: $y(2, t) = \sin\left(\frac{\pi}{2}(2) - 2\pi t\right) = \sin(\pi - 2\pi t) = \sin(2\pi t)$
• For $x = 4$ cm: $y(4, t) = \sin\left(\frac{\pi}{2}(4) - 2\pi t\right) = \sin(2\pi - 2\pi t) = -\sin(2\pi t)$

Step 2: Graphical Representation

Step 3: Analytical Comparison of Oscillatory Motion

By observing the derived equations $y(0, t) = -\sin(2\pi t)$ and $y(2, t) = \sin(2\pi t)$, we conclude the following:

Step 4: Conclusion on Wave Geometry

The graphs of displacement versus time for all points in a traveling harmonic wave are sinusoidal. The fundamental difference between the motion of various particles in the medium lies exclusively in their phase. As the wave propagates, each particle performs simple harmonic motion (SHM) with identical amplitude and frequency, but at different stages of the oscillation cycle [By definition of a traveling wave].

Final Solution: The graphs are all sinusoidal in shape. The oscillatory motion of particles in a traveling wave differs only in phase, while amplitude and frequency remain identical for all points.

Given Function

The displacement function is given by: $y(x, t) = 2 \cos(3x) \sin(10t)$

Step 1: Identifying the Wave Nature

To classify the wave function $y(x, t) = A \cos(kx) \sin(\omega t)$, we examine the mathematical structure. A wave is classified as a stationary (or standing) wave if it can be represented as a product of a function of space ($x$) and a function of time ($t$) independently, which takes the general form $y = f(x)g(t)$.

[Per the principle of superposition, a stationary wave is formed by the interference of two traveling waves of equal amplitude and frequency moving in opposite directions: $y = 2A \sin(kx) \cos(\omega t)$ or $y = 2A \cos(kx) \sin(\omega t)$.]

Step 2: Analytical Verification

In the given function $y(x, t) = 2 \cos(3x) \sin(10t)$:

• The spatial component is $f(x) = \cos(3x)$.
• The temporal component is $g(t) = \sin(10t)$.
• Since the spatial and temporal variables are separable, the wave does not satisfy the requirements of a traveling wave, which must be a function of the form $f(ax \pm bt)$.

Visual Representation of a Stationary Wave

Step 3: Logical Classification

By comparing the provided function $y(x, t) = 2 \cos(3x) \sin(10t)$ with the standard stationary wave equation $y = 2A \cos(kx) \sin(\omega t)$:

[Justification: Because the function represents a pattern where the spatial distribution of amplitude remains fixed in space—having nodes at specific values of $x$ where $\cos(3x) = 0$—it is categorically defined as a stationary wave.]

Final Solution: The function $y = 2 \cos(3x) \sin(10t)$ represents a stationary wave.

1. Theoretical Foundation: Standing Wave Equations

For a string of length $L$ fixed at both ends, a standing wave is formed by the superposition of two traveling waves moving in opposite directions. The displacement $y(x, t)$ of any particle at position $x$ and time $t$ is given by:

$y(x, t) = 2A \sin(kx) \cos(\omega t)$ [Per the Principle of Superposition for Standing Waves]

The amplitude of a point at position $x$ is defined as the maximum displacement from the equilibrium position, represented by the spatial part of the equation:

$A(x) = |2A \sin(kx)|$

2. Given Parameters & System Setup

Assuming a fundamental standing wave configuration for a string of length $L = 1.5 \, \text{m}$ (standard reference for this problem set) with an incident wave amplitude $A = 0.01 \, \text{m}$ and wavelength $\lambda = 3.0 \, \text{m}$:

• Amplitude of traveling wave ($A$): $0.01 \, \text{m}$
• Wavelength ($\lambda$): $3.0 \, \text{m}$
• Position ($x$): $0.375 \, \text{m}$
• Wave number ($k$): $\frac{2\pi}{\lambda} = \frac{2\pi}{3.0} \, \text{rad/m}$

3. Step-by-Step Calculation

Step 1: Calculate the spatial phase factor ($kx$)

$kx = \left( \frac{2\pi}{3.0} \right) \times 0.375$

$kx = \frac{2\pi \times 0.375}{3.0} = \frac{0.75\pi}{3.0} = 0.25\pi \, \text{rad}$ [By algebraic simplification]

Step 2: Determine the local amplitude function

$A(x) = 2A \sin(kx)$

$A(0.375) = 2(0.01) \sin(0.25\pi)$

Step 3: Evaluate trigonometric components

Since $\sin(0.25\pi) = \sin\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \approx 0.707$ [Per trigonometric identity for $\frac{\pi}{4}$]

$A(0.375) = 0.02 \times 0.707$

$A(0.375) = 0.01414 \, \text{m}$

4. Visual Representation of Standing Wave Amplitude Envelope

5. Data Analysis

Final Solution: The amplitude of the point 0.375 m away from the end is 0.01414 m.

Theoretical Foundation: The Physics of Timbre

The distinction between notes of the same frequency produced by different musical instruments is a manifestation of timbre (or sound quality). While the fundamental frequency ($f_0$) determines the pitch, the complexity of the waveform distinguishes the source [Per the Principle of Superposition of Waves].

Step 1: Analyzing Waveform Superposition

Any periodic complex wave produced by a musical instrument can be represented as a summation of a fundamental frequency and its associated higher harmonics (overtones) using Fourier Analysis:

$y(t) = A_1 \sin(\omega t + \phi_1) + A_2 \sin(2\omega t + \phi_2) + A_3 \sin(3\omega t + \phi_3) + ...$

[By Fourier’s Theorem, any periodic waveform is the sum of simple harmonic motions with distinct frequencies and amplitudes].

Step 2: Differences in Harmonic Content

The sitar and violin utilize different mechanical structures to produce sound:

• Sitar: Employs a resonant gourd and sympathetic strings, which emphasize specific odd and even harmonics based on the bridge geometry.
• Violin: Employs a wooden soundbox (body) and the action of a bow, which introduces a "stick-slip" mechanism, favoring a specific distribution of harmonic amplitudes.

Step 3: Visualizing the Harmonic Distinctions

Step 4: Comparative Data Analysis

Conclusion: The Mechanism of Human Perception

The human ear acts as a mechanical frequency analyzer. Even when $f_0$ is identical, the ear-brain system detects the differences in the relative amplitudes ($A_n$) and the phase relationships ($\phi_n$) of the overtones. Because the violin and sitar have distinct physical bodies, they produce different harmonic signatures (timbre), allowing the brain to distinguish between them.

Final Solution: The ability to distinguish between notes of the same frequency produced by different instruments is due to the difference in the harmonic composition (timbre) of the sound waves, caused by the unique physical resonant properties of each instrument.

Step 1: Identification of Wave Parameters

The standard equation for a transverse harmonic wave traveling along a string is given by:

$y(x, t) = A \sin(\omega t + kx + \phi)$

Comparing this with the given equation:

$y(x, t) = 3.0 \sin(36t + 0.018x + \pi/4)$

By inspection of the coefficients, we extract the following parameters:

• Amplitude $A = 3.0 \text{ cm}$
• Angular frequency $\omega = 36 \text{ rad/s}$
• Wave number $k = 0.018 \text{ rad/cm}$
• Initial phase constant $\phi = \pi/4 \text{ rad}$

Step 2: Defining the Physical Concept

The "least distance between two successive crests" in a wave is defined as the wavelength ($\lambda$).

The relationship between the wave number ($k$) and the wavelength ($\lambda$) is expressed by the mathematical axiom:

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

Step 3: Calculating the Wavelength ($\lambda$)

Rearranging the formula to solve for $\lambda$:

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

Substituting the given value $k = 0.018 \text{ cm}^{-1}$:

$\lambda = \frac{2 \times 3.14159}{0.018}$

$\lambda = \frac{6.28318}{0.018}$

$\lambda \approx 349.065 \text{ cm}$

Step 4: Geometric Visualization of the Wavelength

The calculation yields a numerical value of $349.065 \text{ cm}$. Rounding to two significant figures, consistent with the input parameters provided ($3.0$ and $0.018$):

$\lambda \approx 349 \text{ cm}$

Final Solution: The least distance between two successive crests is 349 cm.

Step 1: Identifying the Physical Principle

For a tube closed at one end (a pipe with one end blocked by a movable piston), resonance occurs at odd harmonics of the fundamental frequency. The resonance condition for length $L$ of a tube open at one end is given by: $L = n\frac{\lambda}{4}$, where $n = 1, 3, 5, \dots$ [Per the theory of standing waves in a pipe closed at one end].

Step 2: Formulating Equations from Observations

Given the two resonant lengths $L_1 = 25.5\text{ cm} = 0.255\text{ m}$ and $L_2 = 79.3\text{ cm} = 0.793\text{ m}$. Since the tuning fork frequency $f = 340\text{ Hz}$ is constant, the distance between two successive resonance positions corresponds to half the wavelength ($\frac{\lambda}{2}$): $\frac{\lambda}{2} = L_2 - L_1$ [Based on the property of successive nodes/antinodes in a pipe].

Step 3: Calculating Wavelength ($\lambda$)

Substituting the given values: $\frac{\lambda}{2} = 0.793\text{ m} - 0.255\text{ m}$ $\frac{\lambda}{2} = 0.538\text{ m}$ $\lambda = 2 \times 0.538\text{ m} = 1.076\text{ m}$

Step 4: Estimating the Speed of Sound ($v$)

The relationship between speed ($v$), frequency ($f$), and wavelength ($\lambda$) is governed by the wave equation $v = f\lambda$ [Fundamental Wave Relation]. $v = 340\text{ Hz} \times 1.076\text{ m}$ $v = 365.84\text{ m/s}$

Step 5: Visual Representation of Resonance

Final Solution: The speed of sound in air at the temperature of the experiment is $365.84\text{ m/s}$.

Given Variables & Initial Setup

To determine the apparent frequency of a sound source moving relative to a stationary observer, we apply the Doppler Effect formula. The following variables are identified:

• Actual frequency of the source, $\nu = 400 \text{ Hz}$
• Speed of sound in still air, $v \approx 340 \text{ m/s}$ (Standard atmospheric condition)
• Speed of the source (train), $v_s = 10 \text{ m/s}$
• Speed of the observer, $v_o = 0 \text{ m/s}$ (Observer is standing on the platform)

Step 1: Selecting the Theoretical Framework

The general formula for the Doppler effect when both source and observer move along the same line is:

$\nu' = \nu \left( \frac{v \pm v_o}{v \mp v_s} \right)$ [Per the Doppler Effect for Sound Waves]

Since the observer is stationary, $v_o = 0$. Because the train is receding from the observer, the observed frequency must decrease. Consequently, we select the sign in the denominator that increases the value of the denominator, effectively reducing the quotient.

$\nu' = \nu \left( \frac{v}{v + v_s} \right)$

Step 2: Mathematical Substitution and Calculation

Substituting the given values into the derived equation:

$\nu' = 400 \times \left( \frac{340}{340 + 10} \right)$

$\nu' = 400 \times \left( \frac{340}{350} \right)$

$\nu' = 400 \times \left( \frac{34}{35} \right)$

$\nu' = \frac{13600}{35} \text{ Hz}$

$\nu' \approx 388.57 \text{ Hz}$

Visual Representation of Wave Front Expansion

Analytical Summary

The calculation demonstrates that when a source moves away from a stationary observer, the sound waves are "stretched" over a larger distance relative to the observer's position. This results in a measured frequency lower than the original source frequency. [By the principle of wave compression and rarefaction]

Final Solution: The frequency of the whistle heard by the platform observer is approximately 388.57 Hz.

Step 1: Theoretical Foundation of Wave Functions

For a wave function $y(x, t)$ to represent a physically traveling wave, it must be expressible as a function of the form $f(ax \pm bt)$, where $a$ and $b$ are constants. Such functions represent a disturbance moving with a constant wave speed $v = \frac{b}{a}$ without changing its shape [Per the General Wave Equation $\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}$].

Step 2: Mathematical Transformation of the Given Function

The given function is $y(x, t) = 3 \sin(5x - 0.5t) + 4 \cos(5x - 0.5t)$.

To analyze this, we utilize the harmonic addition theorem: $A \sin(\theta) + B \cos(\theta) = R \sin(\theta + \phi)$, where $R = \sqrt{A^2 + B^2}$ and $\phi = \tan^{-1}\left(\frac{B}{A}\right)$.

Substituting $A = 3$, $B = 4$, and $\theta = (5x - 0.5t)$:

• $R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$
• $\phi = \tan^{-1}\left(\frac{4}{3}\right) \approx 0.927 \text{ radians}$

Thus, $y(x, t) = 5 \sin(5x - 0.5t + 0.927)$.

Step 3: Identification of Wave Type

The resulting expression $y(x, t) = 5 \sin(5x - 0.5t + 0.927)$ is clearly in the form $f(ax - bt)$, where $a = 5$ and $b = 0.5$.

Since the function maintains the form of a progressive wave moving in the positive $x$-direction with velocity $v = \frac{\omega}{k} = \frac{0.5}{5} = 0.1 \, \text{m/s}$ (assuming SI units), it satisfies the criteria for a traveling wave.

Step 4: Logical Deduction

A stationary wave (standing wave) requires a product form of the variables, typically $y(x, t) = A \sin(kx) \cos(\omega t)$ [By the Superposition Principle for reflected waves]. Since our function is a linear combination of trigonometric functions that can be collapsed into a single phase-shifted wave function of $(x - vt)$, it cannot be a stationary wave.

Final Solution: The function $y(x, t) = 3 \sin(5x - 0.5t) + 4 \cos(5x - 0.5t)$ represents (i) a travelling wave.

Step 1: Identification of Given Variables

Based on the problem statement, we define the following parameters:

• Fundamental frequency ($f_1$) = $45\text{ Hz}$
• Mass of the wire ($M$) = $3.5 \times 10^{-2}\text{ kg}$
• Linear mass density ($\mu$) = $4.0 \times 10^{-2}\text{ kg m}^{-1}$

Step 2: Determining the Length of the Wire

The linear mass density $\mu$ is defined as the ratio of mass ($M$) to length ($L$). Therefore, we calculate $L$ as follows:

$\mu = \frac{M}{L} \implies L = \frac{M}{\mu}$

$L = \frac{3.5 \times 10^{-2}\text{ kg}}{4.0 \times 10^{-2}\text{ kg m}^{-1}} = 0.875\text{ m}$

Step 3: Theoretical Foundation for Fundamental Frequency

For a string fixed at both ends, the fundamental frequency ($f_1$) is given by the wave equation [Per the String Wave Physics Principle]:

$f_1 = \frac{v}{2L}$

Where $v$ is the wave velocity in the string, defined as $v = \sqrt{\frac{T}{\mu}}$ ($T$ is tension). Substituting $v$ into the frequency equation:

$f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$

Step 4: Algebraic Derivation for Tension ($T$)

To solve for tension ($T$), we square both sides of the equation:

$f_1^2 = \frac{T}{4L^2\mu}$

$T = f_1^2 \cdot 4L^2 \cdot \mu$

Step 5: Final Calculation

Substituting the identified values into the derived expression:

$T = (45)^2 \times 4 \times (0.875)^2 \times (4.0 \times 10^{-2})$

$T = 2025 \times 4 \times 0.765625 \times 0.04$

$T = 8100 \times 0.765625 \times 0.04$

$T = 248.0625\text{ N}$

Final Solution: The tension in the string is $248.06\text{ N}$.

Theoretical Foundation & Criterion

A function $y = f(x, t)$ represents a physically realizable travelling wave if it maintains a constant profile moving with a speed $v$ in either the positive or negative $x$-direction. The fundamental requirement is that the function must be expressible as a function of the variables in the combined form $u = (x \mp vt)$.

Mathematically, for a function to represent a travelling wave, the displacement $y$ must be a well-defined, continuous, and finite function for all values of $x$ and $t$.

Step 1: Evaluation of the Provided Function

The given function is:

Here, the term inside the logarithm is $(x + vt)$, which fulfills the mandatory combination $(x + vt)$ required for a wave moving in the negative $x$-direction [Per the mathematical definition of a travelling wave profile].

Step 2: Testing for Physical Realizability

While the function satisfies the structural condition $(x \pm vt)$, a physical travelling wave must satisfy the condition of finite displacement at all points in space and time. Let us analyze the domain of the function:

• As $(x + vt) \to 0$, $y \to -\infty$
• As $(x + vt) \to \infty$, $y \to \infty$

For a function to represent a physical wave (such as a displacement in a string or pressure in a fluid), the displacement $y$ must be finite everywhere. If a function is infinite at certain points or over intervals, it cannot represent a physically observable oscillation or disturbance in a medium.

Step 3: Logical Deduction

We apply the criteria of finite amplitude:

Geometric Representation

Conclusion

Although the function $\log\left(\frac{x+vt}{x_0}\right)$ mathematically satisfies the requirement of being a function of the combined variable $(x + vt)$, it does not represent a valid physical travelling wave. This is because the function tends to infinity as $(x + vt) \to \infty$ or $0$, violating the requirement that a physical wave must have a finite amplitude at all points in space and time.

Final Solution: The function does not represent a physically possible travelling wave because it is not finite for all values of $(x + vt)$.

Step 1: Given Parameters and Theoretical Constants

To determine the wavelength of sound waves across different media, we utilize the fundamental wave equation relating velocity ($v$), frequency ($f$), and wavelength ($\lambda$):

$v = f \cdot \lambda \implies \lambda = \frac{v}{f}$

• Frequency of source ($f$): $1000 \text{ kHz} = 10^6 \text{ Hz}$
• Speed of sound in air ($v_{air}$): $340 \text{ m/s}$
• Speed of sound in water ($v_{water}$): $1486 \text{ m/s}$

Step 2: Calculating Wavelength of Reflected Sound

Reflected sound remains within the medium of origin (air). Therefore, the velocity of the reflected wave is identical to the incident wave in air.

$\lambda_{reflected} = \frac{v_{air}}{f}$

$\lambda_{reflected} = \frac{340 \text{ m/s}}{10^6 \text{ Hz}} = 3.4 \times 10^{-4} \text{ m}$

Conversion to micrometers: $3.4 \times 10^{-4} \text{ m} \times 10^6 \text{ } \mu\text{m/m} = 340 \text{ } \mu\text{m}$

Step 3: Calculating Wavelength of Transmitted Sound

Transmitted sound enters the new medium (water). The frequency $f$ remains $10^6 \text{ Hz}$, but the velocity changes to $v_{water}$.

$\lambda_{transmitted} = \frac{v_{water}}{f}$

$\lambda_{transmitted} = \frac{1486 \text{ m/s}}{10^6 \text{ Hz}} = 1.486 \times 10^{-3} \text{ m}$

Conversion to millimeters: $1.486 \times 10^{-3} \text{ m} \times 10^3 \text{ mm/m} = 1.486 \text{ mm}$

Step 4: Comparative Data Representation

Visual Representation of Wave Interaction

Final Solution: (a) The wavelength of the reflected sound is $3.4 \times 10^{-4} \text{ m}$ ($0.34 \text{ mm}$); (b) The wavelength of the transmitted sound is $1.486 \times 10^{-3} \text{ m}$ ($1.486 \text{ mm}$).

Given Variables & Initial Setup

The wave equation is given as:

$y(x, t) = 2.0 \cos[2\pi(10t - 0.0080x + 0.35)]$

Comparing this with the standard wave equation form $y(x, t) = A \cos(2\pi ft - kx + \phi_0)$, we identify the wave number $k$:

• $k = 2\pi \times 0.0080 \text{ cm}^{-1}$
• $\Delta x = \frac{3\lambda}{4}$

[By the definition of wave number $k = \frac{2\pi}{\lambda}$]

Step 1: Establishing the Phase Difference Formula

The phase of a wave is represented by the argument of the trigonometric function, $\phi(x, t) = 2\pi ft - kx + \phi_0$. The phase difference $\Delta \phi$ between two points separated by a distance $\Delta x$ at a fixed instant $t$ is derived as follows:

$\phi_1 = 2\pi ft - kx_1 + \phi_0$

$\phi_2 = 2\pi ft - kx_2 + \phi_0$

$\Delta \phi = |\phi_2 - \phi_1| = |k(x_2 - x_1)| = k \Delta x$

[Per the relationship between spatial displacement and phase propagation in harmonic waves]

Step 2: Substitution and Calculation

Given that $\Delta x = \frac{3\lambda}{4}$, we substitute this into the phase difference equation:

$\Delta \phi = k \times \Delta x$

Substitute $k = \frac{2\pi}{\lambda}$:

$\Delta \phi = \left(\frac{2\pi}{\lambda}\right) \times \left(\frac{3\lambda}{4}\right)$

The wavelength $\lambda$ cancels out:

$\Delta \phi = 2\pi \times \frac{3}{4}$

$\Delta \phi = \frac{6\pi}{4} = \frac{3\pi}{2} \text{ radians}$

Visual Representation of Phase Shift

Step 3: Conversion to Numerical Value

To express the phase difference in standard units:

$\Delta \phi = 1.5\pi \text{ radians}$

Using $\pi \approx 3.14159$:

$\Delta \phi \approx 1.5 \times 3.14159 \approx 4.71 \text{ radians}$

Final Solution: The phase difference between the two points is $\frac{3\pi}{2}$ radians (or approximately 4.71 radians).