Question 214: Waves

Question

A sound absorber attenuates the sound level by $$20$$ $$dB$$. The intensity decreases by a factor of

Accepted Answer

We have, $${L_1} = 10\log \left( {{{{{m I}_1}} \over {{{m I}_0}}}} ight);$$
$${L_2} = 10\,\log \left( {{{{{m I}_2}} \over {{{m I}_0}}}} ight)$$
$$	herefore$$ $$\,\,{L_1} - {L_2} = 10\,\log \left( {{{{{m I}_1}} \over {{{m I}_0}}}} ight) - 10\,\log \left( {{{{{m I}_2}} \over {{{m I}_0}}}} ight)$$
or, $$\Delta L = 10\,\log \left( {{{{{m I}_1}} \over {{{m I}_0}}} 	imes {{{{m I}_0}} \over {{{m I}_2}}}} ight)$$
or, $$\Delta L = 10\,\log \left( {{{{{m I}_1}} \over {{{m I}_2}}}} ight)$$
or, $$20 = 10\log \left( {{{{{m I}_1}} \over {{{m I}_2}}}} ight)$$
or, $$2 = \log \left( {{{{{m I}_1}} \over {{{m I}_2}}}} ight)$$
or, $${{{{m I}_1}} \over {{{m I}_2}}} = {10^2}$$
or, $${{m I}_2} = {{{{m I}_1}} \over {100}}.$$
$$ \Rightarrow $$ Intensity decreases by a factor $$100.$$

Answer

$$100$$

Answer

$$1000$$

Answer

$$10000$$

Answer

$$10$$

Waves

A sound absorber attenuates the sound level by $20 dB$. The intensity decreases by a factor of

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