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# A rock band gives rise to an average sound level of $102\,dB$ at a distance of $20\,m$ from the centre of the band. As an approximation, assuming that the band radiates sound equally into a sphere. The sound power output of the band is $8 \times {10^x}\,watts$. Find $x$.

Last updated date: 02nd Aug 2024
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Hint:We can use the relationship connecting the loudness and the ratio of intensities to find the value of intensity of sound. Using that power can be calculated. Since, power is given as the product of intensity and area. Since it is assumed that sound is radiated in the form of a sphere, we can take the area as the area of the sphere. By comparing the final answer with the answer given in the question we can find the value of x.

Complete step by step solution:
It is given that average sound level is $102\,dB$ at $20m$ distance. That is, loudness is given.
We can denote it as L
$L = 102dB$
It is assumed that the band radiates sound equally into a sphere.
We need to find the power output at a distance of $20m$.
We know that the relationship between loudness and intensity is given as
$L = 10\,{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)$
Where, $I$ is the intensity of sound and ${I_0}$ is the reference intensity.
From this we can get the ratio $\dfrac{I}{{{I_0}}}$ as
$\dfrac{I}{{{I_0}}} = {10^{\left( {\dfrac{L}{{10}}} \right)}}$
On substituting the given values we get,
$\dfrac{I}{{{{10}^{ - 12}}}} = {10^{\left( {\dfrac{{102}}{{10}}} \right)}}$
Since the reference frequency ${I_0}$ is not given we can take it as ${10^{ - 12}}$
$\Rightarrow \dfrac{I}{{{{10}^{ - 12}}}} = {10^{\left( {10.2} \right)}}$
$\Rightarrow I = {10^{ - 12}} \times {10^{\left( {10.2} \right)}}$
$\therefore I = 1 \cdot 59 \times {10^{ - 2}}W/{m^2}$
Power is given by the product intensity $I$ and area $A$ .
That is
$P = I \times A$
We know the area of the sphere is given as
$A = 4\pi {r^2}$
We need to find power at a distance $20\,m$ from the centre of the band. So, we can consider a sphere of radius $20\,m$ .
$\therefore A = 4\pi \times {\left( {20} \right)^2}$
Thus, power is,
$P = 1 \cdot 59 \times {10^{ - 2}} \times 4 \times 3 \cdot 14 \times {\left( {20} \right)^2}\,W$
$\Rightarrow P = 80\,W$
$\therefore P = 8 \times {10^1}W$
It was given that the sound power output of the band is $8 \times {10^x}\,W$.
We need to find the value of x,

On comparing this value with the power that we calculated, we can see that the value of x is 1.

Note: The equation that we used connecting the loudness and the intensities is used in the case when the sound level is given in decibel. The logarithm of the ratio of intensity of sound to reference intensity in $W/{m^2}$ gives us the loudness in the bell. To make it in decibel we need to multiply this term by 10.