Answer: A 10,000-Hz sound is 10 times more intense as compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness.
Explanation:
The formula used is as follows.
[tex]\beta = 10 dB log (\frac{I}{I_{o}})\\60 = 10 dB log (\frac{I}{I_{o}})[/tex]
[tex]I_{o} = 10^{-12}[/tex] normal threshold
The difference is sound level is as follows.
60 - 60 = 0
Hence,
[tex]0 = 10 dB [log (\frac{I_{f}}{I_{o}}) - log (\frac{I_{i}}{I_{o}})]\\log (\frac{1000}{I_{o}}) = log (\frac{10000 x}{I_{o}})\\log (10^{15}) = log (10^{16}x)\\15 = 16 + log x\\log x = 1\\x = 10[/tex]
This means that 10,000 Hz sound is 10 times more intense.
Thus, we can conclude that a 10,000-Hz sound is 10 times more intense as compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness.
