Miller_HW10

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Pennsylvania State University *

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Feb 20, 2024

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AE309 Topic 5a Airborne Sound Insulation: Assignment 10 FIRST SEMESTER | ACADEMIC YEAR 2023 – 2024 | PROF. MICHELLE VIGEANT-HAAS Sound Insulation Questions A) Define transmission coefficient and transmission loss without using equation.  Transmission Coefficient (Symbol: 𝜏 ) o Definition: The transmission coefficient represents the amount of (or power) of energy when it hits a surface and how much passes through. o Unit: Since the coefficient ( 𝜏 ) represents a ratio, then its dimensionless meaning does not have a specific unit. Although it is taken as a percentage.  Transmission loss (Symbol: TL): o Definition: Transmission loss is the amount (or power) that is lost when sound passes through a barrier. o Unit: Transmission loss is typically measured in dB. B) Explain in words what 𝜏 = 1 represents and include 2 examples where this would be the case.  𝜏 = 1 represents a situation where all of the incident energy (100%) is transmitted through a barrier without any loss. The barrier does not absorb, reflect, etc. o Optical Transmission Through Glass or Windows: Ideally, in a case where the glass is perfectly clear with little to no imperfections, 100% of the visible light would pass through. o Radio Frequency Transmission in Vacuum: In the case of radio frequency (RF) they will travel 100% through a vacuum making the transmission coefficient 𝜏 = 1 as none of the medium absorbed any of the (RF). C) Show mathematically Why the tripling the thickness of a single- leaf partition would provide an increase of approximately 9 – 10 dB? Part A  Equations Needed: o TL 1 = 20Log(f) = 20Log( ρL 1 ) – 47 o TL 2 = 20Log(f) = 20Log(# ρL 1 ) – 47 o ΔTL = TL 2 – TL 1  My arithmetic: o 20Log(f) = 20Log(3 ρL 2 ) – 20Log( ρL 1 ) ==> 20Log(3 ρL 2, ρL 1 ) ==> 20Log(3) = 9.542 = 10.0 dB Part B  Mass law modeling is only not valid at low frequencies, below the critical frequencies. o My Drawn example: D) The following video is a demonstration of transmission loss measurements – watch this video starting at 2:10 to the end.  Part A: Fill in the blank: o Maximum Insertion loss  Part B: Brief summary of the measurement procedure. o Date acquisition card is used to connect the microphone to the computer. o National instruments compactDAQ with 9324. o Source is 2 channel loudspeaker. o Receiver is a is model 378B02 – ½” Free-field Condenser Microphone. o The gasket is material is to prevent acoustic leaks and pit it on top of the speaker then add the 5 different panel materials.  1/6 – Inch ABS Plastic  1/8 – Inch Abs Plastic  1/8 – Inch Fiberglass  1/8 – Inch Aluminum Honeycomb  1/4 - Inch Aluminum o Determine the maximum insertion loss playing in the room. o Play sound through the speaker. o Measure the background noise in the room and subtract two levels (device does this for you).  Part C: Complete the table below using answers from d:  Part d: Reference equations are taken from the left-hand side of the page at PART A : Material Thickness (inches) Material Type Surface mass density, m s (kg/m 2 ) Measured Approx. Loss at 3400 Hz (dB) Calculated* Approx. Loss at 3400 Hz (dB) 1 1/16 ABS Plastic 3.5 30 10.42 2 1/8 ABS Plastic 6.0 34.7 21.13 3 1/8 Aluminum Honeycomb 6.0 34/7 15.11 4 1/8 Fiberglass 11.0 39.7 26.40

AE309 Topic 5a Airborne Sound Insulation: Assignment 10 FIRST SEMESTER | ACADEMIC YEAR 2023 – 2024 | PROF. MICHELLE VIGEANT-HAAS  Part e How similar or different: o Material 1 (ABS Plastic, 1/16 inches), the difference is |30 - 10.42| = 19.58 dB. o Material 2 (ABS Plastic, 1/8 inches), the difference is |34.7 - 21.13| = 13.57 dB. o Material 3 (Aluminum Honeycomb, 1/8 inches), the difference is |34.7 - 15.11| = 19.59 dB. o Material 4 (Fiberglass, 1/8 inches), the difference is |39.7 - 26.40| = 13.3 dB.  In all cases, the difference is much greater than ±1 dB suggesting that the theoretical calculation is not providing an accurate prediction of the acoustical measured loss.  Two other possible material properties that could be influencing the acoustic sound performance aside from thickness could be… o Material Density : The mass density of a material can affects sound absorption characteristics. o Dampening Characteristics : Referring to the materials ability to dissipate energy.  Part f: Given the mass law how much should the TL increase by? o The TL should increase by 6dB.  Part g: What two materials in the video are measured by the mass law with an approximate doubling of mass? Is the measured distance consistent with the law? o 1/8” ABS Plastic: The surface mass density of about 6 kg/m 2 . o 1/8” Fiberglass Material: Surface mass density of about 11 kg/m 2 . o In the video it was mentioned that the “actual measured distance” may not always exactly match the mass law predictions due to factors like leaks around the panel mount.  Part h: Complete the table below comparing the loss of a double panel with similar mass.  Part f: Explain why or why not the losses are/aren’t similar despite the mass law predicting the same results. o The measured sound losses tend to vary from the mass law predictions because the law does not account for real-world factors like the amount a panel leaks or the materials properties (i.e., stiffness and internal dampening). In the end, the mass law does not for every property when measuring the material. Description Material Thickness (inches) Material Type Surface mass density, m s (kg/m 2 ) Measured Approx. Loss at 3400 Hz (dB) Single Panel 1/8” Fiberglass 1/8 Fiberglass 11.0 39.7 Double Panel 1/8” Honeycomb + 1/8” ABS Combination of 1/8” panels and with 1/8” gap Honeycomb and ABS 11.6 Greater than 39.7 (exact value unknown)

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