Percentage Change in Sound Intensity - Help

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SUMMARY

An intensity level change of +2.00 dB corresponds to a percentage change in intensity of 58.5%. This calculation is derived from the formula B = dB = 10 log (I/Io), where I represents the final intensity and Io the initial intensity. By substituting 2.00 dB into the equation, the ratio I/Io is determined to be approximately 1.585. The final percentage difference is calculated as 100(1.585 - 1.0).

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  • Understanding of decibel (dB) scale and its applications
  • Familiarity with logarithmic functions
  • Basic knowledge of intensity levels in physics
  • Ability to perform percentage calculations
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  • Learn about logarithmic scales in various scientific contexts
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Students in physics or engineering courses, audio engineers, and anyone interested in understanding sound intensity and its measurement in decibels.

dvolpe
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Homework Statement


An intensity level change of +2.00 dB corresponds to what percentage change in intensity!


Homework Equations



B = dB = 10 log (I/Io)

The Attempt at a Solution


2.00 dB = 10 log (I/Io)
0.2 = log (I/Io)
I/Io = 10e0.2 = 1.585

Assuming Io = 1, then percentage difference = 100(1.585-1.0) = 58.5%.

Is this correct? Please help asap.
 
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This is correct per webassign
 
dvolpe said:
This is correct per webassign
Yup, looks good to me too.
 

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