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Explain the absorption coefficient of a hall. Calculate the change in intensity level when the intensity of sound increases 1000 times its original intensity.

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 5M

Year: Dec 2012

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Absorption Co-efficient:-

• When a sound waves incident on a surface, a part of it gets absorbed by the surface, part of it gets reflected and the rest is transmitted.

• The part of the sound waves absorbed by the surface is given by the Absorption Co-efficient.

• Absorption co-efficient can be defined as the ratio of Energy of sound absorbed by the surface to the Total sound energy incident on the surface

$a = \frac{Energy \ \ of \ \ Sound \ \ absorbed \ \ by \ \ the \ \ surface}{Total \ \ sound \ \ energy \ \ incident \ \ on \ \ the \ \ surface}$

Sound Intensity:-

Let $I_0$ be the original intensity and I be the final intensity

Given that,$I/I_0 = 1000$

Now, the change in intensity (in dB) is given by,

$∆I = 10 log_{10} (I/I_0)$

$∴ ∆I = 10 log_{10} (1000)$

$∴ ∆I = 30dB$

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10views

Absorption Co-efficient:-

• When a sound waves incident on a surface, a part of it gets absorbed by the surface, part of it gets reflected and the rest is transmitted.

• The part of the sound waves absorbed by the surface is given by the Absorption Co-efficient.

• Absorption co-efficient can be defined as the ratio of Energy of sound absorbed by the surface to the Total sound energy incident on the surface

$a = \frac{Energy \ \ of \ \ Sound \ \ absorbed \ \ by \ \ the \ \ surface}{Total \ \ sound \ \ energy \ \ incident \ \ on \ \ the \ \ surface}$

Sound Intensity:-

Let $I_0$ be the original intensity and I be the final intensity

Given that,$I/I_0 = 1000$

Now, the change in intensity (in dB) is given by,

$∆I = 10 log_{10} (I/I_0)$

$∴ ∆I = 10 log_{10} (1000)$

$∴ ∆I = 30dB$

0
12views

Absorption Co-efficient:-

• When a sound waves incident on a surface, a part of it gets absorbed by the surface, part of it gets reflected and the rest is transmitted.

• The part of the sound waves absorbed by the surface is given by the Absorption Co-efficient.

• Absorption co-efficient can be defined as the ratio of Energy of sound absorbed by the surface to the Total sound energy incident on the surface

$a = \frac{Energy \ \ of \ \ Sound \ \ absorbed \ \ by \ \ the \ \ surface}{Total \ \ sound \ \ energy \ \ incident \ \ on \ \ the \ \ surface}$

Sound Intensity:-

Let $I_0$ be the original intensity and I be the final intensity

Given that,$I/I_0 = 1000$

Now, the change in intensity (in dB) is given by,

$∆I = 10 log_{10} (I/I_0)$

$∴ ∆I = 10 log_{10} (1000)$

$∴ ∆I = 30dB$

0
7views

Absorption Co-efficient:-

• When a sound waves incident on a surface, a part of it gets absorbed by the surface, part of it gets reflected and the rest is transmitted.

• The part of the sound waves absorbed by the surface is given by the Absorption Co-efficient.

• Absorption co-efficient can be defined as the ratio of Energy of sound absorbed by the surface to the Total sound energy incident on the surface

$a = \frac{Energy \ \ of \ \ Sound \ \ absorbed \ \ by \ \ the \ \ surface}{Total \ \ sound \ \ energy \ \ incident \ \ on \ \ the \ \ surface}$

Sound Intensity:-

Let $I_0$ be the original intensity and I be the final intensity

Given that,$I/I_0 = 1000$

Now, the change in intensity (in dB) is given by,

$∆I = 10 log_{10} (I/I_0)$

$∴ ∆I = 10 log_{10} (1000)$

$∴ ∆I = 30dB$