Answer:
The amplitude of wave increases by a factor of 20
Explanation:
The Sound level in decibel β is described as
[tex]\beta =(10dB)log(\frac{I}{I_{o} })[/tex]
Where I₀=10⁻¹²W/m² is reference intensity
Suppose β₁ is initial sound intensity and β₂ is final sound intensity.Therefore the change in sound intensity is given as
β₁-β₂=26.0 dB
[tex](10dB)log(\frac{I_{2} }{I_{o}} )-(10dB)log(\frac{I_{1} }{I_{o}} )=26.0dB\\(10dB)[log(\frac{I_{2} }{I_{o}} )-log(\frac{I_{1} }{I_{o}} )]=26.0dB\\log(\frac{\frac{I_{2}}{I_{o}} }{\frac{I_{1}}{I_{o}} })=\frac{26dB}{10dB}\\ log\frac{I_{2}}{I_{1}}=2.6\\ \frac{I_{2}}{I_{1}}=10^{2.6}[/tex]
The intensity of sound is directly proportional to the amplitude of wave squared
So
I∝A²
[tex]\frac{I_{1}}{I_{2}}=(\frac{A_{1}}{A_{2}})^{2}\\ (\frac{A_{1}}{A_{2}})=\sqrt{(\frac{I_{1}}{I_{2}})}\\ (\frac{A_{2}}{A_{1}})=\sqrt{(\frac{I_{2}}{I_{1}})}\\ (\frac{A_{2}}{A_{1}})=\sqrt{10^{2.6} }\\ (\frac{A_{2}}{A_{1}})=20[/tex]
Therefore the amplitude of wave increases by a factor of 20
