Acoustic Materials And Meta Materials

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| By Anguru Kishore
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Anguru Kishore
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Quizzes Created: 1 | Total Attempts: 77
Questions: 30 | Attempts: 77

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Acoustic Materials And Meta Materials - Quiz


Questions and Answers
  • 1. 

    In a travelling wave, the oscillation of individual particles of the medium

    • A.

      A). Vary with space but not with time.

    • B.

      B). Vary with time but not with space.

    • C.

      C). Vary with the both time and space.

    • D.

      D). All the above.

    Correct Answer
    C. C). Vary with the both time and space.
    Explanation
    In a travelling wave, the oscillation of individual particles of the medium varies with both time and space. This means that as the wave propagates through the medium, the particles not only oscillate back and forth in time, but also vary in their displacement from their equilibrium position as a function of their position in space. This is characteristic of a travelling wave, where the disturbance is transferred from one particle to another, causing them to oscillate in a coordinated manner. Therefore, option c) is the correct answer.

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  • 2. 

    In standing wave, the individual particles of the medium

    • A.

      A). Oscillate at fixed amplitude not vary with time, vary only with the space.

    • B.

      B). Oscillate at fixed amplitude not vary with Space, vary only with the time.

    • C.

      C). Oscillate at fixed amplitude varies with space and time.

    • D.

      D). None of the above.

    Correct Answer
    B. B). Oscillate at fixed amplitude not vary with Space, vary only with the time.
    Explanation
    In a standing wave, the individual particles of the medium oscillate at a fixed amplitude, meaning their displacement from the equilibrium position remains constant. However, the variation in their position occurs only with respect to time, as the particles move back and forth in a periodic manner. This is because the standing wave is formed by the interference of two waves traveling in opposite directions, causing the particles to oscillate in place rather than propagate through space. Therefore, option b) is the correct explanation.

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  • 3. 

    In a harmonic plane wave, where the acoustic variable

    • A.

      A). has constant amplitude on any plane perpendicular to the direction of wave propagation.

    • B.

      B). has constant phase on any plane perpendicular to the direction of wave propagation.  

    • C.

      C). has constant amplitude and phase on any plane perpendicular to the direction of wave propagation.

    • D.

      D). All the above.

    Correct Answer
    C. C). has constant amplitude and pHase on any plane perpendicular to the direction of wave propagation.
    Explanation
    In a harmonic plane wave, the acoustic variable refers to the quantity that is being measured or observed, such as pressure or displacement. The statement "has constant amplitude and phase on any plane perpendicular to the direction of wave propagation" means that the magnitude and timing of the acoustic variable remain the same on any plane that is perpendicular to the direction in which the wave is traveling. This suggests that the wave is uniform and does not change in intensity or timing as it propagates through space. Therefore, option c) is the correct answer.

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  • 4. 

    What is a wave number

    • A.

      A). Is the number of complete wave cycles in radians per unit time.

    • B.

      B). Is the number of complete wave cycles in radians per unit spatial dimension.

    • C.

      C). Is the number of complete wave cycles in radians per sec.

    • D.

      D). None of the above.

    Correct Answer
    B. B). Is the number of complete wave cycles in radians per unit spatial dimension.
    Explanation
    The wave number is a measure of the number of complete wave cycles in radians per unit spatial dimension. It represents the spatial frequency of a wave, indicating how rapidly the wave oscillates within a given distance. This measurement is important in various fields, such as physics and engineering, where understanding the characteristics and behavior of waves is crucial. The correct answer, option b, accurately describes the concept of wave number.

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  • 5. 

    Reflection coefficient of an acoustic material is the

    • A.

      A). ratio of complex pressure amplitudes between the transmitted and incident waves.

    • B.

      B). lost of incident energy in the process of reflection.

    • C.

      C). ratio of complex pressure amplitudes between the reflected and incident waves.

    • D.

      D). Both b and c.

    Correct Answer
    C. C). ratio of complex pressure amplitudes between the reflected and incident waves.
    Explanation
    The reflection coefficient of an acoustic material is a measure of the ratio of complex pressure amplitudes between the reflected and incident waves. It quantifies how much of the incident wave is reflected back from the material. This coefficient is important in understanding the behavior of sound waves when they encounter different materials, as it helps determine the amount of reflection that occurs. Option c correctly describes the reflection coefficient in terms of the ratio between the reflected and incident waves' complex pressure amplitudes.

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  • 6. 

    What happens when sound wave propagating in one medium encounters the boundary of a second  medium?

    • A.

      A). The wave is get diffracted.

    • B.

      B). Partly get reflected.

    • C.

      C). Partly get transmitted.

    • D.

      D). All the above.

    Correct Answer
    D. D). All the above.
    Explanation
    When a sound wave propagating in one medium encounters the boundary of a second medium, multiple things can happen. Firstly, the wave can undergo diffraction, which is the bending of the wave around obstacles or through openings. Secondly, a portion of the wave can get reflected, meaning it bounces back from the boundary. Lastly, another portion of the wave can get transmitted, meaning it passes through the boundary and continues propagating in the second medium. Therefore, all of these options are correct.

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  • 7. 

    The Eigen frequencies of a constrained medium are given by:

    • A.

      A). fn=c/2L

    • B.

      B). fn=nc/4L

    • C.

      C). fn=nc/2L

    • D.

      D). fn=nc/L

    Correct Answer
    C. C). fn=nc/2L
    Explanation
    The correct answer is c). fn=nc/2L. This equation represents the eigen frequencies of a constrained medium. The variable n represents the mode number, c represents the speed of propagation in the medium, and L represents the length of the medium. The equation shows that the eigen frequencies are directly proportional to the mode number and the speed of propagation, and inversely proportional to the length of the medium.

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  • 8. 

    Logarithmic or dB scales are used for measuring

    • A.

      A). Sound pressure level

    • B.

      B). Sound intensity level

    • C.

      C). Sound power level

    • D.

      D). All the above

    Correct Answer
    D. D). All the above
    Explanation
    Logarithmic or dB scales are used for measuring sound pressure level, sound intensity level, and sound power level. These scales are used because they allow for a more convenient representation of the wide range of values that can be associated with these measurements. The logarithmic scale compresses the values, making it easier to compare and analyze different levels of sound. Additionally, the use of dB scales allows for the addition and subtraction of sound levels, which is important in various applications such as noise control and audio engineering.

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  • 9. 

    The minimum threshold of hearing of human ear is

    • A.

      A). 1x10-5 Pa .

    • B.

      B). 2x10-5 Pa.

    • C.

      C). 20 Pa .

    • D.

      D). 200 Pa.

    Correct Answer
    B. B). 2x10-5 Pa.
    Explanation
    The minimum threshold of hearing of the human ear is the lowest sound intensity that can be detected by the ear. It is commonly accepted that the threshold of hearing is around 0 dB SPL, which corresponds to a sound pressure level of 2x10-5 Pa. This means that any sound with a lower intensity than this value would not be audible to the human ear. Therefore, the correct answer is b). 2x10-5 Pa.

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  • 10. 

    Human hearing studies show that the two sounds are distinguished in terms of

    • A.

      A). their pressure difference

    • B.

      B). the ratio of their intensities

    • C.

      C). both a & b

    • D.

      D). their absolute pressures

    Correct Answer
    B. B). the ratio of their intensities
    Explanation
    Human hearing is able to distinguish between two sounds based on the ratio of their intensities. This means that the difference in loudness or volume between the two sounds is what allows us to perceive them as distinct. The absolute pressures of the sounds do not play a role in this distinction.

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  • 11. 

    What is the increase in dB scale for every increment of pressure by 10 times

    • A.

      A). 1dB

    • B.

      B). 5dB

    • C.

      C). 10dB

    • D.

      D). 20dB

    Correct Answer
    D. D). 20dB
    Explanation
    The dB scale is a logarithmic scale used to measure the intensity or magnitude of a sound or pressure. It is based on a ratio of the pressure being measured to a reference pressure. In this case, the question states that the pressure is being incremented by 10 times. Since the dB scale is logarithmic, an increase of 10 times in pressure corresponds to a 20dB increase on the scale. Therefore, the correct answer is d) 20dB.

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  • 12. 

    Sound pressure level is defined as

    • A.

      A). Lp = 10 log10 (prms /Pref)

    • B.

      B). Lp = 10 log10 (Irms/Iref)

    • C.

      C). Lp = 20 log10 (prms /Pref)

    • D.

      D). Both b and c

    Correct Answer
    D. D). Both b and c
    Explanation
    The correct answer is d) Both b and c. Both options b) and c) are correct because they represent different ways of calculating sound pressure level. Option b) calculates sound pressure level using the root mean square of the sound pressure (Irms) divided by the reference sound pressure (Iref), while option c) calculates sound pressure level using the root mean square of the sound pressure (prms) divided by the reference sound pressure (Pref), multiplied by 20. Both equations are valid and can be used to calculate sound pressure level.

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  • 13. 

    Based on subjective perception of sounds, the minimum sound level that is considered to be noisy is

    • A.

      A). 60dB

    • B.

      B). 80dB

    • C.

      C). 100dB

    • D.

      D). 140dB

    Correct Answer
    B. B). 80dB
    Explanation
    The minimum sound level that is considered to be noisy is subjective and can vary from person to person. However, on average, 80dB is often considered to be the threshold for noise. At this level, prolonged exposure can cause damage to hearing.

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  • 14. 

    A Random noise with equal energy per  frequency  and thus  constant spectral density level over all frequencies is called

    • A.

      A). Blue noise

    • B.

      B). Pink noise

    • C.

      C). White noise

    • D.

      D). All the above

    Correct Answer
    C. C). White noise
    Explanation
    White noise is a random noise that has equal energy per frequency and a constant spectral density level over all frequencies. It is called white noise because it contains all frequencies in equal amounts, similar to white light which contains all colors in equal amounts. Blue noise and pink noise are specific types of noise with different spectral characteristics, so they are not correct answers. Therefore, the correct answer is c) White noise.

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  • 15. 

    What kind of noise is used for masking purpose and especially in speech tests

    • A.

      A). Blue noise

    • B.

      B). Pink noise

    • C.

      C). White noise

    • D.

      D). All the above

    Correct Answer
    C. C). White noise
    Explanation
    White noise is used for masking purposes and especially in speech tests. White noise is a type of noise that contains all frequencies in equal amounts. It is often used to mask or cover up other sounds, making them less audible. In speech tests, white noise can be used to create a consistent background noise, allowing for accurate measurement of speech perception and detection thresholds. Blue noise and pink noise are also types of noise, but they are not specifically used for masking purposes in speech tests.

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  • 16. 

    Two sounds are said to be coherent if they have

    • A.

      A). same frequency

    • B.

      B). same wave form

    • C.

      C). constant phase difference

    • D.

      D). All the above

    Correct Answer
    D. D). All the above
    Explanation
    Coherent sounds have the same frequency, meaning they have the same number of cycles per second. They also have the same wave form, which refers to the shape of the wave that represents the sound. Additionally, coherent sounds have a constant phase difference, which means that the two waves are always in sync with each other. Therefore, all of the given options (a, b, and c) are correct explanations for coherent sounds.

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  • 17. 

    The SPL is measured by a

    • A.

      A). Amplifier

    • B.

      B). Signal conditioner

    • C.

      C). Sound level meter

    • D.

      D). Signal Analyser

    Correct Answer
    C. C). Sound level meter
    Explanation
    The correct answer is c) Sound level meter. The SPL (Sound Pressure Level) is a measure of the intensity of sound. It is typically measured using a sound level meter, which is a device specifically designed to measure and quantify sound levels. Amplifiers, signal conditioners, and signal analyzers are not specifically designed for measuring sound levels and may not provide accurate or reliable measurements.

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  • 18. 

    The frequency bands that used for quick frequency analysis is

    • A.

      A). octave bands

    • B.

      B). 1/3 octave bands

    • C.

      C). 1/6 octave bands

    • D.

      D). Any one of the above.

    Correct Answer
    A. A). octave bands
    Explanation
    Octave bands are commonly used for quick frequency analysis because they divide the frequency spectrum into octave intervals, which are logarithmic and evenly spaced. This allows for a simplified and efficient analysis of the frequency content of a signal or sound. 1/3 octave bands and 1/6 octave bands provide more detailed frequency analysis but are not typically used for quick analysis due to their narrower bandwidths. Therefore, the correct answer is octave bands.

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  • 19. 

    The centre frequency of an octave band is

    • A.

      A). Difference of upper and lower frequencies

    • B.

      B). Average of upper and lower frequencies

    • C.

      C). Geometric mean of upper and lower frequencies

    • D.

      D). None of the above

    Correct Answer
    C. C). Geometric mean of upper and lower frequencies
    Explanation
    The centre frequency of an octave band is calculated by taking the geometric mean of the upper and lower frequencies. The geometric mean is the square root of the product of the upper and lower frequencies. This method is used to determine the central frequency of an octave band because it provides a logarithmic scale that evenly divides the frequency range.

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  • 20. 

    What is the probably best and cost effective strategy for noise control is

    • A.

      A). Source based

    • B.

      B). Path based

    • C.

      C). Receiver based

    • D.

      D). Any one of the above

    Correct Answer
    A. A). Source based
    Explanation
    Source based noise control is the most effective and cost-effective strategy for noise control. This approach focuses on reducing or eliminating the noise at its source, such as using quieter machinery or equipment, implementing soundproofing measures, or modifying the design of the source to minimize noise production. By addressing the source of the noise, it prevents the noise from propagating through the path or reaching the receiver, making it a more efficient and economical solution compared to path or receiver-based strategies.

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  • 21. 

    Which one of the following statement is correct

    • A.

      A). Acoustic materials are designed to reduce noise by controlling, directing and Manipulating sound waves.

    • B.

      B). Noise control with acoustic materials is typically achieved by controlling the reflection and transmission through these materials.

    • C.

      C). Whenever a sound wave propagating in some fluid medium suddenly if it encounters a boundary then a part of energy is reflected, some part gets transmitted.

    • D.

      D). All the above

    Correct Answer
    D. D). All the above
    Explanation
    All the above statements are correct. Acoustic materials are indeed designed to reduce noise by controlling, directing, and manipulating sound waves. Noise control with these materials is typically achieved by controlling the reflection and transmission through them. Additionally, when a sound wave encounters a boundary in a fluid medium, a part of its energy is reflected and some part gets transmitted. Therefore, all the statements in options a), b), and c) are correct.

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  • 22. 

    An acoustic field where the sound waves reflect numerous times and reach the observer from all directions is called

    • A.

      A). Near field

    • B.

      B). Far field

    • C.

      C). Diffusive filed

    • D.

      D). Open field

    Correct Answer
    C. C). Diffusive filed
    Explanation
    An acoustic field where the sound waves reflect numerous times and reach the observer from all directions is called a diffusive field. In this type of field, the sound waves scatter and spread out in all directions, creating a diffuse sound field. This is in contrast to the near field, where the sound waves are close to the source and have not had a chance to spread out, and the far field, where the sound waves have spread out and become more uniform. An open field refers to an outdoor environment without any obstructions.

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  • 23. 

    Which one of the following statement is incorrect

    • A.

      A). Insertion loss of a material is the difference in sound pressure levels at the receiver point without and with the material.

    • B.

      B). Nosie reduction is the difference between the sound pressure level before and after passing through a barrier material.

    • C.

      C). Transmission loss is the difference between the incident sound intensity level and transmitted sound intensity level when the sound waves hit that material.

    • D.

      D). None of the above.

    Correct Answer
    D. D). None of the above.
    Explanation
    The correct answer is d) None of the above. This means that all of the statements given in options a), b), and c) are correct. Option a) states that insertion loss is the difference in sound pressure levels with and without a material, which is true. Option b) states that noise reduction is the difference in sound pressure levels before and after passing through a barrier material, which is also true. Option c) states that transmission loss is the difference between incident and transmitted sound intensity levels when sound waves hit a material, which is again true. Therefore, the correct answer is d) None of the above.

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  • 24. 

    What is absorption coefficient

    • A.

      A). the ratio of sound intensity absorbed by the material to the sound intensity incident on the material.

    • B.

      B). the energy absorbed is the fraction of the incident energy that is lost in the process of reflection

    • C.

      C). lost of incident energy in the process of reflection.

    • D.

      D). All the above

    Correct Answer
    D. D). All the above
    Explanation
    The absorption coefficient is a measure of how much sound energy is absorbed by a material. It is the ratio of the sound intensity absorbed by the material to the sound intensity incident on the material. In other words, it represents the fraction of the incident energy that is lost in the process of reflection. Therefore, all of the given options are correct explanations of the absorption coefficient.

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  • 25. 

     If a noise control is desired mainly in medium1 along with medium2, then the acoustic material should

    • A.

      A). Reflect less

    • B.

      B). Transmit less

    • C.

      C). Dissipate more

    • D.

      D). All the above

    Correct Answer
    D. D). All the above
    Explanation
    If a noise control is desired mainly in medium1 along with medium2, then the acoustic material should reflect less, transmit less, and dissipate more. This means that the material should not bounce back or reflect the noise, should not allow the noise to pass through it easily, and should absorb or dissipate the noise energy. By doing all of these, the material will effectively control the noise in both medium1 and medium2.

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  • 26. 

    Performance of an enclosures

    • A.

      A). Decrease in transmission coefficient of the enclosures wall materials

    • B.

      B). Increase with absorption coefficient of the lining material

    • C.

      C). Increase with wall material thickness

    • D.

      D). All the above

    Correct Answer
    D. D). All the above
    Explanation
    The performance of an enclosure is determined by multiple factors. The transmission coefficient of the enclosure's wall materials affects how much sound is transmitted through the walls. A decrease in this coefficient would result in better performance. Additionally, the absorption coefficient of the lining material inside the enclosure also plays a role. A higher absorption coefficient means more sound energy is absorbed, improving the enclosure's performance. Lastly, the thickness of the wall materials can also impact performance. Thicker walls provide better sound insulation. Therefore, all of the options mentioned (a, b, and c) contribute to the overall performance of an enclosure.

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  • 27. 

    Which one of the following statement is incorrect

    • A.

      A). Opaque/semi opaque materials are used for acoustic barriers

    • B.

      B). Barriers for outdoor is mostly use in the free field conditions

    • C.

      C). Barriers are much effective in direct sound field compared to the reflected sound file.

    • D.

      D). None of the above

    Correct Answer
    D. D). None of the above
    Explanation
    The correct answer is d) None of the above. This means that all of the statements provided in options a), b), and c) are correct. Opaque/semi-opaque materials are indeed used for acoustic barriers, barriers for outdoor use are mostly used in free field conditions, and barriers are more effective in the direct sound field compared to the reflected sound field.

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  • 28. 

    Dissipation mechanism of porous materials in porous sound absorbers is due to

    • A.

      A). Viscous shear

    • B.

      B). Friction

    • C.

      C). Scattering

    • D.

      D). All the above

    Correct Answer
    D. D). All the above
    Explanation
    The dissipation mechanism of porous materials in porous sound absorbers is due to a combination of viscous shear, friction, and scattering. Viscous shear occurs when sound waves pass through the small pores in the material, causing the air particles to rub against each other and convert the sound energy into heat. Friction also plays a role in dissipating sound energy as it causes air particles to collide with the porous material, converting kinetic energy into heat. Additionally, scattering occurs when sound waves interact with the irregular surfaces of the porous material, causing the sound energy to be dispersed in different directions and reducing its intensity. Therefore, all of these mechanisms contribute to the dissipation of sound energy in porous sound absorbers.

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  • 29. 

    What are all the different types of sound absorbing materials:

    • A.

      A). Porous-fibrous sound absorbers

    • B.

      B). Panel sound absorbers

    • C.

      C). Helmholtz resonators

    • D.

      D). All the above.

    Correct Answer
    D. D). All the above.
    Explanation
    The correct answer is d) All the above. This means that all the different types of sound absorbing materials mentioned in options a, b, and c are correct. Porous-fibrous sound absorbers, panel sound absorbers, and Helmholtz resonators are all types of materials that can absorb sound.

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  • 30. 

    Low cost, light weight, eco friendly solution for broad band noise control at high frequencies is

    • A.

      A). Barriers

    • B.

      B). Enclosures

    • C.

      C). Porous absorbers

    • D.

      D). Helmholtz resonators

    Correct Answer
    C. C). Porous absorbers
    Explanation
    Porous absorbers are a low-cost, lightweight, and eco-friendly solution for broadband noise control at high frequencies. They are made of porous materials that absorb sound energy by converting it into heat. These absorbers are effective in reducing noise in various applications such as industrial settings, transportation, and architectural spaces. They are preferred over barriers and enclosures because they are easier to install, require less space, and provide better absorption of high-frequency noise. Helmholtz resonators, on the other hand, are used for specific frequency ranges and may not be as effective for broadband noise control.

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Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.

  • Current Version
  • Mar 20, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • May 22, 2020
    Quiz Created by
    Anguru Kishore
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