Physics, asked by sachinidesilva772, 6 months ago

An intensity level change of 1dB corresponds to what percentage change in intensity?

Answers

Answered by raaziyatasleem
2

Answer: can't help I am in small class

Answered by mad210215
0

1db intensity level change corresponds to change in intensity of 12.2%

Explanation:

Given : L_{p} = 1 dB

where  L_{p} = intensity pressure level in decibels

To find : %Δp where p = intensity level

Formulae :

L_{p} = 20log_{10}\frac{p}{p_{0} }

%Δp = %Δp = \frac{(p-p_{0} }{p_{0} }\frac{p-p_{0} }{p_{0} } x 100

where

p = new intensity level

p_{0} = old intensity level

Solution :

Step-1: Substitute given values in formula L_{p} = 20log_{10}\frac{p}{p_{0} } => 1 = 20xlog_{10}\frac{p}{p_{0} }

Step-2 : Rearrange values of equation from Step-1 : log_{10}\frac{p}{p_{0} } = 1/20 = 0.05

Step-3 : Take antilog for the equation of Step-2 : \frac{p}{p_{0} } = antilog (0.05) = 1.122

Step-4 : Subtract 1 from both sides of equation in Step-3 :

\frac{p}{p_{0} } - 1 = 1.122 - 1 = 0.122

Step-5 : Simplify and multiply by 100 on both sides of equation in Step-4 :  \frac{p-p_{0} }{p_{0} }  x 100 = 0.122 x 100 = 12.2

Step-6 : Apply formula %Δp = %Δp = \frac{(p-p_{0} }{p_{0} }\frac{p-p_{0} }{p_{0} } x 100 in Step-6 to get : %Δp = 12.2%

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