• The notes detail various concepts and formulas related to acoustics, music, and wave phenomena

Identifying Combinations of Frequencies

• Listener is exposed to two frequencies (_ Hz and _ Hz) in the right ear and one frequency (_ Hz) in the left ear
• Listener determines the combination through finding the greatest common divisor (GCD) of the frequencies to identify the corresponding tone

Fourier Components of a Complex Waveform

• Complex waveform repeats every _ ms
• Possible Fourier component frequencies (in Hz) are determined by: Fundamental frequency = 1 / Period (P) = _ Hz, correct answer includes multiples of the fundamental frequency

Sound Intensity & Sound Intensity Level (SIL) at a Distance

• Foghorn's sound intensity is measured at _ W/m² at _ m
• To find intensity/SIL at _ km, use Inverse Square Law:
  • ²= ² → I₂=I₁(₁)² (r₁: initial distance, r₂: doubled distance, I₁: measured intensity _ W/m²)
• Numerator: intensity at measurement point (furthest point, decibel level)
• Denominator: base intensity at starting point with total intensity being l₀
• When calculating the decibel difference, use the starting point intensity as a reference
• Calculate SIL using: SIL(dB) = 10log₁₀ ( ) → I₀ = 10 −12 W/m²

Interference from 2 Loudspeakers

• Standing midway between two loudspeakers (one on each side) with musical notes causing constructive interference
• To achieve destructive interference, need to move left
  • Moving ¼ wavelength closer to one speaker results in moving ¼ wavelength farther from the other speaker, creating a ½ wavelength difference
• Destructive interference occurs when the path difference is ½ wavelength
• Constructive interference occurs when the path difference is an integer wavelength

Coupled Pendula w/ a Stiffer Spring

• Standing between two loudspeakers with musical notes causing constructive interference
• Altering the connecting spring between two coupled pendula impacts frequencies
  • Mode 1 (Lower freq): Coupled Pendula move together causing the spring to have no effect and frequencies remain unchanged
  • Mode 2 (Higher freq): Coupled Pendula move opposite which causes the spring to stretch/compress and increases the restoring force which increases the frequency

Piano Tuning and Frequency Deviation

• Hitting a key on the piano while a tuner app is showing a value higher than the standard equal temperament tuning (A₄=440 Hz)
• The fundamental frequency of the tone played is calculated by: f = standard tuning frequency * 10^(C/3986)

Tremolo Definition

• Modulation/variation in tone's loudness and sound level

Fourier Components of a Note

• A complex waveform producing a note
• Possible Fourier components are the fundamental note frequencies and multiples of fundamental frequencies

Effect of Doubling Tension on Piano String

• Doubling tension on a _ piano string affects the resulting tone
  • Frequency is proportional to the square root of tension with f ~ √T

Effect of Temperature on Sound Parameters

• Temperature variations affect only frequency and speed
• Speed of sound is temperature dependent with initiating vibrations unchanged
• Consequence of affecting speed of sound is that only frequency and speed are affected and not amplitude or wavelengths

Short Answer

• Uniform narrow tube _ m long, open at both ends, resonates at successive harmonics _ Hz and _ Hz:
  • What is the fundamental frequency? fbeats = |f₁- f₂|
  • Speed of sound in gas: f₁ = → 𝑣 = 2𝐿f₁ (open); f₁ = → 𝑣 = 4𝐿f₁ (closed)
• If tube is closed at one end, the difference in frequencies of two consecutive harmonics: fbeats = 2(|𝑓₁ − 𝑓₂|)
• Second (open) tube sounded with the original tube produces a beat frequency of _ Hz
  • What length might the second tube have?
• The second tube resonates at the fundamental +/- the new beat-frequency produced _ Hz → f₁ = → L = 𝑣
  • Plug in speed found in part (b) for v, and compute 2 lengths, one for each frequency found by adding or subtracting the new frequency from the fundamental

Combination Tones

• 3 primary tones form the major chord _,, and
• Additional notes that may arise as difference tones which are found by:
  • Determine tone frequency
  • Subtract all possible combinations like | f₁ - f₂|, |f₁- f₃|, |f₂ - f₃|
  • Match resultant frequencies to their corresponding tones for difference, summation, and cubic difference tones

Fundamentals of Sound Waves & Acoustics

• Velocity (in m/s)
  • V20 (T = 20°C): 344 m/s or 1130 ft/s or 770 mph
  • VT temp is equal to 344 + 0.6(T – 20°) m/s with every °C rise in air temperature above 20°C and the speed of sound increases by 0.6 m/s
  • v = displacement over time and v = fx
• Distance (in m): d = 4a, d = V1V2 = V2V1
• Pressure (in N/m²): P = Force over Area with S being measured in units N/m
  • Pressure in compressions is P = P + ΔΡ
  • ∆P= Pc-P, with Pc measured during compression (atm), Po is normal atm (atm), ΔΡ is change in pressure from equilibrium, and Pl pressure during refraction (atm)
• Wavelength (in m) = f/𝑣 and Frequency (in Hz): f = λ over𝑣 or f = 𝜆
• Period (in s): T = 1/𝑓
• Amplitude (in m): Wave amplitude in pascals (N/m²): Apa = A x 10⁵ , A (in amplitude in N/m²) with 1 atm = 10² N/m²
• Sound Waves Travel Through Different Mediums
  • d air is airVelocity*airTime
  • d concrete is concreteVelocity*concreteTime
  • Timesound difference = (Concrete time - Air Time)
  • Concrete TIme = (sound Difference)/(airVelocity−concreteVelocity)
  • Air time = (sound Difference)/(airVelocity−concreteVelocity)
  • Medium dependence measured at V₂₀: Helium = 100 m/s; CO₂ (pure) = 270 m/s; H₂O = 6000 m/s; Steel = 6000 m/s; Speed of sound in concrete: 3000m/s
  • "5 second rule": for every 5 sec between seeing lightning and hearing thunder, the lightning is 1 mile away
  • Percent Error: % error = [actual distance - estimated distance]/ actual distance * 100
  • Angular Frequency = = 2πf
  • Human Hearing: range is 20 Hz - 20,000 Hz
  • Soft Sound vs. Loud Sound: Amplitude in the range of a few Pascals is soft sound while an amplitude of several hundred Pascals or more is loud sound

Examples

• How far will sound travel in a specific time?
  • At T = 20°C, d = t * v → d = t * 344m/s
  • At T = _°C, d = t * v → d = t * (344 + 0.6(T – 20°)m/s)
  • Error made over one mile using the "5-second rule" with the temperature at _°C:
  • Calculate speed of sound: 344 + 0.6(_°C - 20°) m/s Calculate the actual distance: d = v * t → (speed of sound (in m/s)) (5s) = meters % error = actual distance –estimated distance/actual Distance * 100 % error = (((344 +0.6(T-20°) m/s)(5s)] - estimated distance)/ [((344 +0.6(T-20°) m/s)(5s)] x 100% d estimate distance: 1 mile = 1610 meters
• On a warm summer day (°C₁), it takes t₁ for an echo to return from a cliff across a lake
  • to winter °C₂ in order to calculate echo? d = V₁t₁ =V₂t₂→ d = [344 + 0.6( v₁− 20°) m/s](t₁) = [344 + 0.6(v₁− 20°) m/s](t₂) and that t₁ is only given

Simple Harmonic Motion (SHM)

• Period (in s): Т = 2П√(m/k)
• Frequency (in Hz): f = (1/2π) √(k/m)
  • mf² = k/ 4π² and m₁f₁² = m₂f₂²
• Angular Frequency is (in Hz): ω = √(k/m)
• Restoring Force: Hooke's Law (in N): F_spring = kx, x = displacement from equilibrium
• Spring Constant (in N/m) and Stiffness are also measured by k = F_ext/x
  • 4π² m/𝑇², = 3mg/Δ𝑥, 1 N/m = 1 kg/s², also Displacement (in m): Ax = mg divided by k
• Displacement (in m) can be Δx = Acos(ωt+ ϕ), A = amplitude and ϕ = phase angle
• Potential Energy (in J): PE = ½kx² max @ max displacement (x = ± A)) and Kinetic Energy (in J): KE = ½mv² max @ equilibrium (x = 0))
• Total Mechanical Energy (in J): E = PE + KE = ½kA²
• Work (in J): W = FD, (D = distance, F = force) Acceleration of Gravity (m/s²): g = 9.8 m/s²

Examples

• Find frequency of sound wave whose vibrations occur _ ms apart f = (1/ P)+x 1000 = in Hz (1000 ms = 1 s)
• Find amplitude of two waves: (highest crest of wave - lowest crest of wave) = m
• If an object undergoes SHM w/ amplitude _ m, the total distance it travels in one period? d = 4a (in m) mass m at the end of a spring
• Frequency Oscillates with 𝑓1, and When an additional 𝑚 mass is added to m, the frequency is f₂:
  • Measured by the value of m: (m kg)(f₁)² = (m kg + 𝑚1𝑘𝑔 )(f₂)² → 𝑚 = (𝑚1𝑓2)²/((𝑓2)² - (𝑓₁)²)

Wave Interference, Diffraction, Doppler Effect

• Beats (in Hz): 𝑓 beats = |𝑓₁ - 𝑓₂ |

• of Cycles: 𝑛 = 𝑡Τ−¹ → t = time, T = period of wave

• Length difference L1-L2 = n ^-1→ n = number of cycles
• Diffraction (in m):
  • D ≤λ for strong diffraction with D = size of obstacle
  • Constructive Interference: L1- L2 = n λ(n = integer 0, 1, 2...)
  • Destructive Interference: L1- L2 = (n + ½) λ

Doppler Effect

• Frequency shifts with sound frequency change based on motion
• When speaker moves the frequency increases and When moving away the decreases
• If two speakers are the same frequencies and move at same rate their frequencies stay equal, resulting in a beat rate of 0 If person between frequencies and move, the frequencies differ creating at beat
• Beat rate only sounds when different frequency

Doppler-Examples

• The observer would observe frequency from the moving tuba of: f = stationary frequency/1 = (approaching speed/speed of sound )
• The Drop is 12% is f_new = f₀ ((1+v_s /v )=v_old ((1 − v_s /v ) Vs The person ₁₁-relative velocity of the source and Observer V is equal to the Speed f=( )f₀ (V: relative speed of the source and observer, v: speed of signal, of the Signal which only applies to when a sound source is moving f'=f(() v₀ = Speed of Observer is positive of moving source vs= Speed of force is positive of moving source In a pillar the The Lower Freq is higher Wavelength is better which is better than blocks The shorter wavelength over 1200 will be blocked

Example of a Tuba Experiment for beat frequency and Doppler Effect

• Station observer's would observe by: f′=(1+−approaching rate/ speed of sound) ⋅ the stationary frequency = f₀(1 +15"345. )=85.9 Hz The beat frequency is the subtraction of 88.9 Hx to 85 Hz which means the total beat for f = 3.9Hz

Musician reporting Drop by 12% of the frequency

New pitch is 12%, so f2=1.12 f₁ is v0speed+vs/speed of V=Vspedd-vr Intensity

Sound is within Power/Area

• Signal Intensity, SIL1-sil2 =10 dl SL (DB0=(10 log i)/(10 log i₀) =10 to log (I/IO based of decal, intensity over IO and intensity=()^2 Range=√ intensity ratios Reference intensity/Decibal level : IO= 10^12-SIL=0

Surface Area vs Sound intensity Formula

• Surface area is 4pir^4
• Sound intensity= P\ 4pir^4 or = POwers/(st)=(Intensity 1)
• Inverse Square Law=(11)/(r 1)^2=(Intensity/ Intensity2)/(r_2 )^2 Power\ Energy P=IS vs energy PE^t Area Relationship=(Area^2)/(Intensity^4(1+A)/(a^2) The relation : Constructive Amplitude A1=A2=4AIntensity1=16I Destructive Amplitude =1/A(A1=A2IntensityI= AI (1+1/2A1+2(A+A)) The loudness, intensity I=E\ST is measured in with with =SY I=0 Estimate at cm with intensity assume = W1 ^2 And spreads roughly in sound =I and intensity is P 4pi power

Example of Firework over sound in Joules

The Explosion=(log (area)/( log (area^2+height ^2)10(3/4)

Basic Music Overview

• Harmonic Series: f= nf²
• Decibal Scale: SL DB= 10 LOG L- LOG Io or change is LOG (P)-logo for 2 signals

Tempo range is Largo Presto

Fast-120-or m4- meaning for core note to 120perminute typical ragen 1=50 large-200 presto Four core notebats per measure 404/2226-eight notebats per meter typically measured in 2 sets ForTempo Formula BPM divided by 6= second speed Frequency is the same (2^(7 \ power 12 )) 1^ Power to semitone 12 = 1.5 941 4 Key-in=between with two notes= semitomes but 2 Yite beats one note 1 Semitones Basic Frequency =n*fn

Expression Range

• Expression, Smooth and Stacott which affects an articulation with muddiness Find Frequencies to each with Equal Tempered Chronmatic Scale- Count semitones for with formula is frequency 15 Power to 22 to With 1,3, 4 what is Harmonies series Flatter or sharper low with high % with what is a set If You go to and 4 what is the 11 How many What is P1 ratio what is C ratio Tuning and music

Pythagorean &Musical Range

Has Mathematical Imperfection Circle OF moving though and original Chromatic is used for P1 with divide note by

Intervals

• Intervals of notes go what 1/P2
• What is a series of Sound- Amplitude of signal is Sine 471 741 1 is open is 4 is close T= Simple form or Squred to reduce or to Sharpness and harmnoses equal with low in Music 1/F212= and F /F /31=3 with for and

4Interval Units

Interval 120012 -19 - with with What the frequency of that is the result what from P=F F= From to Find what frequency with that Find 1 st 8 harmonics

Examples

• The Note by Complex waveform What frequency = P_ = frequency +F= Find Firsts Hamonic by What Frequency the missed Hamonic
• Missing fundamental Match missing Find Harmonics state the frequencies a+ Harmonics - How would resonant 21 is deep How with the

Human Hearing Overview

Human range =Young vs Old Sensitivity= but increase Is pitch the same with a great Pitch and what is sound 2 Sign wave can Wave= accuracy =J=20 to 10 is more decrease In A*