Answer
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Hint: Sound Intensity, also known as acoustic intensity, is the power the sound wave carries per unit area in a direction perpendicular to the aforementioned area. Decibel, on the other hand, is a logarithmic unit, used to measure sound level.
Formula used: \[\beta =10\log \dfrac{I}{{{I}_{0}}}\]
Complete step by step solution:
We have been given that intensity of sound increases by a factor of \[30\]
One decibel is equal to ten times the logarithm to base \[10\] (or common logarithm) of the power or the intensity ratio. It can be more clearly expressed as a formula,
\[\beta =10\log \dfrac{I}{{{I}_{0}}}\] where \[\beta \] is the sound level in decibels, \[I\] is the intensity of sound and \[{{I}_{0}}\] is the threshold intensity of sound.
Let the initial intensity of the sound be \[I\], we can express it in decibels as \[{{\beta }_{1}}=10\log \dfrac{I}{{{I}_{0}}}\]
Now, the intensity of sound is increased by a factor of \[30\], so the new intensity of the sound will be \[30I\]. The loudness of this intensity can be expressed as \[{{\beta }_{2}}=10\log \dfrac{30I}{{{I}_{0}}}\]
Since we are concerned with the increase in the loudness, we can find it by taking the difference between the two calculated decibel loudness,
Increase in sound level \[\Rightarrow {{\beta }_{2}}-{{\beta }_{1}}\]
\[{{\beta }_{2}}-{{\beta }_{1}}=10\log \dfrac{30I}{{{I}_{0}}}-10\log \dfrac{I}{{{I}_{0}}}\]
Using properties of logarithms, we can now say that
\[\begin{align}
& {{\beta }_{2}}-{{\beta }_{1}}=10(\log \dfrac{30(\dfrac{I}{{{I}_{0}}})}{1(\dfrac{I}{{{I}_{0}}})}) \\
& \Rightarrow {{\beta }_{2}}-{{\beta }_{1}}=10\log 30=14.77dB \\
\end{align}\]
Hence, there is an increase of \[14.77dB\] in the sound level when intensity increases by a factor of \[30\].
Note:Loudness refers to how loud or soft a sound seems to a listener. The loudness of sound is determined by its intensity and intensity, in turn, is determined by the amplitude of the sound waves and the distance travelled by the sound waves from the source.
Formula used: \[\beta =10\log \dfrac{I}{{{I}_{0}}}\]
Complete step by step solution:
We have been given that intensity of sound increases by a factor of \[30\]
One decibel is equal to ten times the logarithm to base \[10\] (or common logarithm) of the power or the intensity ratio. It can be more clearly expressed as a formula,
\[\beta =10\log \dfrac{I}{{{I}_{0}}}\] where \[\beta \] is the sound level in decibels, \[I\] is the intensity of sound and \[{{I}_{0}}\] is the threshold intensity of sound.
Let the initial intensity of the sound be \[I\], we can express it in decibels as \[{{\beta }_{1}}=10\log \dfrac{I}{{{I}_{0}}}\]
Now, the intensity of sound is increased by a factor of \[30\], so the new intensity of the sound will be \[30I\]. The loudness of this intensity can be expressed as \[{{\beta }_{2}}=10\log \dfrac{30I}{{{I}_{0}}}\]
Since we are concerned with the increase in the loudness, we can find it by taking the difference between the two calculated decibel loudness,
Increase in sound level \[\Rightarrow {{\beta }_{2}}-{{\beta }_{1}}\]
\[{{\beta }_{2}}-{{\beta }_{1}}=10\log \dfrac{30I}{{{I}_{0}}}-10\log \dfrac{I}{{{I}_{0}}}\]
Using properties of logarithms, we can now say that
\[\begin{align}
& {{\beta }_{2}}-{{\beta }_{1}}=10(\log \dfrac{30(\dfrac{I}{{{I}_{0}}})}{1(\dfrac{I}{{{I}_{0}}})}) \\
& \Rightarrow {{\beta }_{2}}-{{\beta }_{1}}=10\log 30=14.77dB \\
\end{align}\]
Hence, there is an increase of \[14.77dB\] in the sound level when intensity increases by a factor of \[30\].
Note:Loudness refers to how loud or soft a sound seems to a listener. The loudness of sound is determined by its intensity and intensity, in turn, is determined by the amplitude of the sound waves and the distance travelled by the sound waves from the source.
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