Amplitude of sound wave in by 50%

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SUMMARY

The amplitude of a sound wave increased by 50% results in a corresponding increase in the decibel level of approximately 1.8 dB. This calculation is based on the formula β = 10 log(I/I₀), where I represents the intensity of the sound wave. Using initial values of I = 2.0 x 10^-7 and I₀ = 1.00 x 10^-12, the initial decibel level is calculated to be 53 dB. After increasing the intensity by 50%, the new decibel level is approximately 54.8 dB, confirming the increase of 1.8 dB.

PREREQUISITES
  • Understanding of sound wave properties
  • Familiarity with the decibel scale
  • Knowledge of logarithmic calculations
  • Basic concepts of intensity in acoustics
NEXT STEPS
  • Study the relationship between amplitude and intensity in sound waves
  • Learn about the logarithmic scale in acoustics
  • Explore the implications of decibel changes in audio engineering
  • Investigate the effects of sound wave amplitude on human perception
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Acoustics students, audio engineers, sound technicians, and anyone interested in understanding sound wave behavior and decibel calculations.

hemetite
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Any kind soul to help me with this questions...i am stuck

qn. The amplitude of vibration of a certain sound wave is increase by 50%. What is the corresponding increase in the decibel level of the sound?

My attempt..

beta= 10 log ( I/Io)

lets just take I = 2.0 x 10^-7
let just take Io= 1.00 x 10^-12

then 10 log( 2.0 x 10^-7/1.00 x 10^-12) = 10 log (2.0 x 10^5) = 53db

if we increse I by 50 % = (2.0 x 10^-7) x 1.5
= 3.0 x 10^-7

then
10 log( 3.0 x 10^-7/1.00 x 10^-12) = 10 log (3.0 x 10^5) = 54.8 db

so the change of decibel = 54.8db - 53db = 1.8db

correct attemp this question this way??
 
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hemetite said:
Any kind soul to help me with this questions...i am stuck

qn. The amplitude of vibration of a certain sound wave is increase by 50%. What is the corresponding increase in the decibel level of the sound?

My attempt..

beta= 10 log ( I/Io)

lets just take I = 2.0 x 10^-7
let just take Io= 1.00 x 10^-12

then 10 log( 2.0 x 10^-7/1.00 x 10^-12) = 10 log (2.0 x 10^5) = 53db

if we increse I by 50 % = (2.0 x 10^-7) x 1.5
= 3.0 x 10^-7

then
10 log( 3.0 x 10^-7/1.00 x 10^-12) = 10 log (3.0 x 10^5) = 54.8 db

so the change of decibel = 54.8db - 53db = 1.8db

correct attemp this question this way??

Since the increase is by a factor of 50% you can also express the increase
as simply 10log10(1.5) = 10*(0.176) = 1.76 db
 

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