Comprehensive Study Notes on Physics of Sound and Waves

Kinematics and Speed of Sound Calculations

• Average Speed and Unit Conversions          Average speed is defined as total distance traveled divided by total elapsed time:          $v_{avg} = \frac{d}{t}$

  • Single-Leg Trip Calculation:                  For a car traveling from Mechanicsburg, PA to Carlisle, PA ($17\,km$) in $23\,min$:                  $17\,km = 17,000\,m$$23\,min = 23 \times 60\,s = 1,380\,s$$v_{avg} = \frac{17,000\,m}{1,380\,s} = 12.3\,m/s$
  • Round-Trip Calculation:                  For the entire round trip where the return journey takes $32\,min$ due to traffic:                  $\text{Total Distance} = 17\,km + 17\,km = 34\,km = 34,000\,m$$\text{Total Time} = 23\,min + 32\,min = 55\,min = 3,300\,s$$v_{avg} = \frac{34,000\,m}{3,300\,s} = 10.3\,m/s$

• Propagation Time of Nuclear Test Shockwaves          During nuclear testing conducted at Bikina Atoll between 1946 and 1958, a monitoring buoy positioned $10\,km$ ($10,000\,m$) away detects shockwaves transmitted through different media:     

  • Detection in Seawater:                  The speed of sound in seawater is $1533\,m/s$.                  $t = \frac{10,000\,m}{1,533\,m/s} = 6.5\,s$
  • Detection in Air at $20^\circ\text{C}$:                  The speed of sound in air at $20^\circ\text{C}$ is $343\,m/s$.                  $t = \frac{10,000\,m}{343\,m/s} = 29.2\,s$

• Ultrasound Echo Distance and Ambient Temperature Determination          An ultrasound detector emits a pulse and receives an echo $2\,ms$ ($0.002\,s$) later reflected off a wall $35.2\,cm$ away.     

  • Determining Sound Velocity:                  Because the wave travels to the wall and back, the total distance traveled is twice the wall distance:                  $d_{total} = 2 \times 35.2\,cm = 70.4\,cm = 0.704\,m$$v_s = \frac{70.4\,cm}{0.002\,s} = 35,200\,cm/s = 352\,m/s$
  • Calculating Room Temperature:                  The speed of sound in air as a function of Celsius temperature ($T_c$) is given by:                  $v_s = 331 \sqrt{1 + \frac{T_c}{273}}$                  Substituting $352\,m/s$ into the formula and solving for $T_c$:                  $352 = 331 \sqrt{1 + \frac{T_c}{273}}$$\left(\frac{352}{331}\right)^2 = 1 + \frac{T_c}{273}$$T_c = 273 \left[ \left(\frac{352}{331}\right)^2 - 1 \right]$$T_c = 35.7^\circ\text{C}$

Pressure, Force, and Area Relationships

• Fundamental Pressure Formula          Pressure ($P$) is defined as force ($F$) exerted per unit area ($A$):          $P = \frac{F}{A} \implies F = P A$

• Single-Point vs. Flat Pressure Calculations          A man standing flat on one foot with a surface area of $31\,\text{sq. in}$ exerts a floor pressure of $4.8\,\text{lbs/sq. in}$.     

  • Total Body Weight/Force:                  $F = P A = (31\,\text{sq. in}) \times (4.8\,\text{lbs/sq. in}) = 148.8\,\text{lbs}$
  • Pressure Exerted by a Single Nail:                  If the same force is concentrated on a single nail head of area $0.4\,\text{sq. in}$:                  $P_{nail} = \frac{148.8\,\text{lbs}}{0.4\,\text{sq. in}} = 372\,\frac{\text{lbs}}{\text{sq. in}}$

• Physics of the Bed of Nails (Dr. Farrar Scenario)          Dr. Farrar can stand comfortably on a bed of 100 nails, but cannot stand on a single nail. The physical mechanisms underlying this are:     

  • The downward force applied by Dr. Farrar is identical in both scenarios and equals his body weight.
  • Because pressure is force per unit area, 100 nails provide 100 times the surface contact area of a single nail.
  • Consequently, the pressure exerted at each individual nail tip on the bed of nails is 100 times less than the pressure exerted by a single nail.

• Pressure Amplitude Variation with Distance in Spherical Waves          In acoustic propagation, sound pressure amplitude $P$ varies inversely with distance $r$ from a point source:          $\frac{P_2}{P_1} = \frac{r_1}{r_2}$          Given a loudspeaker output measuring $12\,kPa$ at a distance of $4.3\,m$, the pressure $P_{10}$ at a distance of $10\,m$ is calculated as:          $\frac{P_{10}}{12\,kPa} = \frac{4.3\,m}{10\,m}$$P_{10} = 5.16\,kPa$

Wave Mechanics and Properties

• Nature and Classification of Waves

  • All waves represent the propagation of energy through space and time.
  • Sound waves propagate through a physical medium (such as air) via alternating compression and decompression of particles. Compressions represent localized regions of elevated potential energy, whereas rarefactions correspond to regions of elevated kinetic energy. This pressure wave is capable of displacing objects and performing physical work.
  • Ocean waves are classified as transverse waves, wherein medium particle movement (up-and-down water displacement) occurs perpendicular to wave propagation direction (toward the shore). Thus, a floating boat oscillates vertically without moving closer to or farther from the shore.

• Particle Velocity vs. Wave Propagation Velocity

  • The speed of sound does not equal the speed of individual air molecules.
  • Air molecules jitter back and forth with a broad thermal distribution of speeds.
  • Sound waves represent the overall propagation velocity of a localized pressure disturbance through the medium, not the transport velocity of physical air molecules.

• Analysis of Pressure-Time Waveforms

              Evaluating a sinusoidal wave plot of pressure disturbance $\Delta P\,(\text{atm})$ versus time $(\text{s})$:      * Amplitude: The peak deviation from equilibrium, $A = 2\,\text{atm}$. * Period ($T$): The duration required for one full wave cycle to complete, $T = 0.25\,s$. * Frequency ($f$):                  $f = \frac{1}{T} = \frac{1}{0.25\,s} = 4\,Hz$          * Wavelength ($\lambda$) Calculation at $10^\circ\text{C}$:          1. Determine speed of sound in air at $10^\circ\text{C}$:                          $v_s = 331 \sqrt{1 + \frac{10}{273\,K}} = 337\,m/s$              2. Calculate wavelength via wave equation $v = \lambda f$:                          $337\,m/s = 4\,Hz \times \lambda$$\lambda = \frac{337}{4} = 84\,m$

• Pure Tones, Period, and Pitch Perception

          * The pitch of a sound wave is governed directly by its frequency ($f$). * Period ($T$) and frequency ($f$) are inversely proportional:                  $f = \frac{1}{T}$          * When comparing Tone A and Tone B: Tone B completes full cycles in less time, meaning it possesses a shorter period ($T$), higher frequency ($f$), and consequently a higher pitch.

Sound Intensity, Loudness, and Perception

• Intensity at Ear Opening          For a total sound power of $\mathcal{P} = 1.00 \times 10^{-5}\,W$ delivered to an ear canal opening of radius $r = 0.5\,cm = 0.005\,m$ at a distance of $3\,m$:          $A_{ear} = \pi r^2 = \pi (0.005\,m)^2$$I_{3m} = \frac{\mathcal{P}}{A_{ear}} = \frac{1.00 \times 10^{-5}\,W}{\pi (0.005\,m)^2} = 0.1273\,\frac{W}{m^2}$

• The Decibel Scale          Sound intensity level ($I_{dB}$) in decibels is logarithmic:          $I_{dB} = 10 \log\left(\frac{I}{I_0}\right)$          where $I_0 = 1.0 \times 10^{-12}\,W/m^2$ represents the threshold of human hearing.     

  • Converting $0.1273\,W/m^2$ to decibels:                  $I_{dB} = 10 \log\left(\frac{0.1273\,W/m^2}{1.0 \times 10^{-12}\,W/m^2}\right) = 111\,dB$

• Inverse Square Law for Sound Intensity          Sound intensity emitted isotropically by a point source obeys an inverse-square relationship with radial distance ($r$):          $I = \frac{\mathcal{P}_{source}}{4\pi r^2}$

  • Comparing intensity at $1\,m$ ($I_{1m}$) versus $3\,m$ ($I_{3m}$):                  $\frac{I_{1m}}{I_{3m}} = \frac{\frac{\mathcal{P}_{source}}{4\pi (1)^2}}{\frac{\mathcal{P}_{source}}{4\pi (3)^2}} = \frac{3^2}{1^2} = 9$$I_{1m} = 9 I_{3m} = 9 \times 0.1273\,W/m^2 = 1.14\,\frac{W}{m^2}$

• Energy Ratios across Spherical Surfaces

  • Total Energy Ratio: As a wave propagates outward, total energy conserved across expanding concentric spheres remains constant. The ratio of total sound energy at $10\,m$ compared to $4.3\,m$ is $1$.
  • Energy Per Unit Area Ratio: Energy density per area decreases proportionally to sphere surface area ($4\pi r^2$):                  $\frac{E_{10}}{E_{4.3}} = \frac{A_{4.3}}{A_{10}} = \frac{4\pi (4.3)^2}{4\pi (10)^2} = 0.18$

• Decibel Calculation for Multiple Sound Sources          To determine the combined decibel level of 3 bluetooth speakers when 1 speaker plays at $40\,dB$:     

  1. Convert $40\,dB$ to absolute intensity ($I$) in $W/m^2$:                  $40\,dB = 10 \log\left(\frac{I}{1.0 \times 10^{-12}\,W/m^2}\right)$$4 = \log\left(\frac{I}{1.0 \times 10^{-12}\,W/m^2}\right)$$10^4 = \frac{I}{1.0 \times 10^{-12}\,W/m^2} \implies I = 1.0 \times 10^{-8}\,W/m^2$
  2. Sum the sound intensities of 3 identical speakers:                  $I_{total} = 3 \times 1.0 \times 10^{-8}\,W/m^2 = 3.0 \times 10^{-8}\,W/m^2$
  3. Convert combined intensity back into decibels:                  $I_{dB} = 10 \log\left(\frac{3.0 \times 10^{-8}\,W/m^2}{1.0 \times 10^{-12}\,W/m^2}\right) = 44.8\,dB$

• Fletcher-Munson Equal Loudness Curves (Phon Scale)

  • By definition, $1\,phon$ matches the perceived loudness of $1\,dB$ SPL at a standard reference frequency of $1000\,Hz$. A $30\,dB$ tone at $1000\,Hz$ defines the $30\,phon$ line.
  • Frequency Determination: A pure tone measuring $60\,dB$ that sounds equal in loudness to $30\,dB$ at $1000\,Hz$ ($30\,phon$) corresponds to a frequency of $80\,Hz$ on equal-loudness contours.
  • Intensity Determination: A $100\text{-}Hz$ pure tone sounding equal in loudness to a $20\,dB$ tone at $1000\,Hz$ ($20\,phon$) requires an physical intensity level of approximately $50\,dB$ at $100\,Hz$.

Wave Interference, Beats, and Phase

• Acoustic Interference Mechanics          Two sound tones played simultaneously at equal volumes do not automatically sound twice as loud as a single tone due to wave interference:     

  • Close Frequencies (Within a few Hz): Tones periodically switch between constructive and destructive alignment, creating rhythmic volume pulsing called beats at a characteristic beat frequency.
  • Identical Frequencies: Waves interfere standardly based on spatial path difference, causing standing regions of purely constructive or destructive interference.
  • Widely Disparate Frequencies: Interference patterns cycle too rapidly to produce distinct steady-state phase cancelations; human perception integrates them such that combined output sounds roughly twice as loud.

• Beat Frequency Calculations          Beat frequency equals absolute frequency difference:          $f_{beat} = |f_1 - f_2|$          When one speaker plays at $f_1 = 3000\,Hz$ and produces $f_{beat} = 2\,Hz$:          $|3000 - f_2| = 2 \implies f_2 = 3002\,Hz \quad \text{or} \quad f_2 = 2998\,Hz$

  • Maximum possible frequency: $3002\,Hz$.
  • Minimum possible frequency: $2998\,Hz$.

• Phase Difference and Wave Superposition          Combining two waves of identical frequency and amplitude $A$ produces a resultant wave of amplitude $A_{tot} = \frac{3}{2}A$.     

  • Phase Angle ($\phi$) Derivation:                  $A_{tot} = 2A \cos(\phi/2)$$\frac{3}{2} A = 2A \cos(\phi/2)$$\frac{3}{4} = \cos(\phi/2)$$\frac{\phi}{2} = \cos^{-1}\left(\frac{3}{4}\right) = 41.4^\circ$$\phi = 81.8^\circ$
  • Resultant Intensity Calculation:                  Using wave superposition intensity formula with base intensity $I_0 = 1\,W/m^2$:                  $I = 4 I_0 \cos^2(\phi/2) = 4(1) \cos^2(81.8^\circ / 2) = 2.25 I_0$                  Alternatively, because intensity is proportional to amplitude squared ($I \propto A^2$):                  $I = \left(\frac{3}{2}\right)^2 (1\,W/m^2) = \frac{9}{4}(1) = 2.25\,\frac{W}{m^2}$

• Two-Source Spatial Interference (Speakers and Microphones Setup)

              Consider two speakers separated horizontally by $30\,cm$ emitting sound at velocity $v = 343\,m/s$:      * Constructive Interference at Microphone #1:                  Microphone #1 is positioned along the axis $70\,cm$ in front of Speaker 2.          * Path difference = distance between speakers = $30\,cm = 0.3\,m$. (The absolute distance of $70\,cm$ to Microphone #1 is unnecessary, as interference depends exclusively on path difference). * Condition for constructive interference:                          $\text{Path Difference} = n \lambda_n, \quad n = 1, 2, 3, \dots$$\lambda_n = \frac{\text{Path Difference}}{n}$$\lambda_1 = 30\,cm = 0.3\,m, \quad \lambda_2 = 15\,cm = 0.15\,m, \quad \lambda_3 = 10\,cm = 0.10\,m$              * Frequency relationship $f_n = \frac{v}{\lambda_n}$:                          $f_1 = \frac{343}{0.3} = 1143\,Hz$$f_2 = \frac{343}{0.15} = 2287\,Hz$$f_3 = \frac{343}{0.10} = 3430\,Hz$              * Lowest frequency for constructive interference at Microphone #1 is $1143\,Hz$.          * Destructive Interference at Microphone #2:                  Microphone #2 is positioned $60\,cm$ perpendicular to Speaker 2.          * Path distance from Speaker 2 = $60\,cm$. * Path distance from Speaker 1 (using Pythagorean theorem) = $\sqrt{30^2 + 60^2}\,cm$. * Path difference = $\sqrt{30^2 + 60^2} - 60 = 7.08\,cm = 0.0708\,m$. * Condition for destructive interference:                          $\text{Path Difference} = n \frac{\lambda_n}{2}, \quad n = 1, 3, 5, \dots$$\lambda_n = 2 \times \frac{\text{Path Difference}}{n}$$\lambda_1 = \frac{2(7.08)}{1} = 14.2\,cm = 0.142\,m$$\lambda_2 = \frac{2(7.08)}{3} = 4.72\,cm = 0.0472\,m$$\lambda_3 = \frac{2(7.08)}{5} = 2.83\,cm = 0.0283\,m$              * Frequencies ($f_n = \frac{v}{\lambda_n}$):                          $f_1 = \frac{343}{0.142} = 2,415\,Hz$$f_2 = \frac{343}{0.0472} = 7,266\,Hz$$f_3 = \frac{343}{0.0283} = 12,120\,Hz$              * Lowest frequency for destructive interference at Microphone #2 is $2,415\,Hz$.