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

Question

A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of (A) 1000 (B) 10000 (C) 10 (D) 100

Tardigrade Admin · Accepted Answer

Let intensity of sound be I and I prime. Loudness of sound initially β1=10 log ((I/I0)) Later, β2=10 log ((I prime/I0)) Given, β2-β1=20 ∴ 20 =10 log ((I prime/I)) I prime =100 I

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

1000

10000

10

100

Let intensity of sound be $I$ and $I^{'}$ .
Loudness of sound initially
$β_{1} = 10 lo g (\frac{I}{I _{0}})$
Later, $β_{2} = 10 lo g (\frac{I ^{'}}{I _{0}})$
Given, $β_{2} - β_{1} = 20$
$∴ 20 = 10 lo g (\frac{I ^{'}}{I})$
$I^{'} = 100 I$

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

1000

10000

10

100

Let intensity of sound be I and I′. Loudness of sound initially β1​=10log(I0​I​) Later, β2​=10log(I0​I′​) Given, β2​−β1​=20 ∴20=10log(II′​) I′=100I

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

Let intensity of sound be $I$ and $I^{'}$ .
Loudness of sound initially
$β_{1} = 10 lo g (\frac{I}{I _{0}})$
Later, $β_{2} = 10 lo g (\frac{I ^{'}}{I _{0}})$
Given, $β_{2} - β_{1} = 20$
$∴ 20 = 10 lo g (\frac{I ^{'}}{I})$
$I^{'} = 100 I$