Question

Spherical sound waves are emitted uniformly in all directions from a point source. The variation in sound level SL as a function of distance 'r' from the source can be written as
$\begin{aligned} & \text{A}\text{. }SL=-b\log {{r}^{a}} \\\ & \text{B}\text{. }SL=a-b{{\left( \log r \right)}^{2}} \\\ & \text{C}\text{. }SL=a-b\log r \\\ & \text{D}\text{. }SL=a-\dfrac{b}{{{r}^{2}}} \\\ \end{aligned}$

Answer 1

We are given a point source from which spherical sound waves are emitted uniformly in all directions. To find the variation in sound level we know the equation for sound level. By applying the intensity of sound from a point source at a distance in the equation and converting the equation as a function of r, we will get the solution.
Formula used:
Sound level, $SL=10{{\log }_{10}}\left( \dfrac{I}{{{I}_{0}}} \right)$
Intensity of sound at a distance r, $I=\dfrac{{{k}^{2}}}{r}$

Complete answer:
In the question it is said that there are spherical sound waves emitting uniformly in all directions from a point source.
We know that the intensity of sound from a point source at a distance are is,
$I=\dfrac{{{k}^{2}}}{r}$, where ‘k’ is a constant and ‘r’ is the distance at which the intensity is felt.
Now we need to find the variation in sound level from the source.
To find the sound level we have an equation,
$SL=10{{\log }_{10}}\left( \dfrac{I}{{{I}_{0}}} \right)$. were ‘SL’ is sound level, ‘I’ is the intensity at a distance ‘r’ and ‘${{I}_{0}}$’ is the intensity of the sound wave.
In the above equation, we can substitute the value of ‘I’
This we get,
$\Rightarrow SL=10{{\log }_{10}}\left( \dfrac{k}{{{r}^{2}}{{I}_{0}}} \right)$
We know that $\log \left( \dfrac{a}{b} \right)=\log a-\log b$. Let us substitute this in the above equation.
$\Rightarrow SL=10{{\log }_{10}}k-10{{\log }_{10}}\left( {{r}^{2}}{{I}_{0}} \right)$
We also know that $\log \left( {{a}^{2}}b \right)=2\log a+\log b$.
Hence we can write $10{{\log }_{10}}\left( {{r}^{2}}{{I}_{0}} \right)$ as $2\times 10{{\log }_{10}}\left( r \right)+10{{\log }_{10}}\left( {{I}_{0}} \right)$
Therefore we get sound level as,
$\Rightarrow SL=\Rightarrow 10{{\log }_{10}}k-\left( 2\times 10{{\log }_{10}}\left( r \right)+10{{\log }_{10}}\left( {{I}_{0}} \right) \right)$
$\Rightarrow SL=10{{\log }_{10}}k-2\times 10{{\log }_{10}}\left( r \right)-10{{\log }_{10}}\left( {{I}_{0}} \right)$
We need the variation in sound level as a function of distance ‘r’ from the source.
For that first let us rearrange the equation as,
$\Rightarrow SL=10{{\log }_{10}}k-10{{\log }_{10}}\left( {{I}_{0}} \right)-2\times 10{{\log }_{10}}\left( r \right)$
Now let us consider,
$10{{\log }_{10}}k-10{{\log }_{10}}\left( {{I}_{0}} \right)=a$ and
$2\times 10{{\log }_{10}}=b$
Therefore we get sound level as,
$\Rightarrow SL=a-b{{\log }_{10}}\left( r \right)$

So, the correct answer is “Option C”.

Note:
Intensity of sound is defined as the energy carried by a wave over some period of time per unit area. The area will be always perpendicular to the flow of energy.
Intensity of a wave will always be directly proportional to the square of amplitude of the wave.
If we have a sound wave with the reference intensity equal to the intensity of the sound wave, then the sound level of that wave will be zero, i.e.
If $I={{I}_{0}}$
We will get sound level,
$SL=10{{\log }_{10}}\left( \dfrac{{{I}_{0}}}{{{I}_{0}}} \right)$
$\Rightarrow SL=10{{\log }_{10}}\left( 1 \right)$
We know that ${{\log }_{10}}\left( 1 \right)=0$
Therefore,
$\Rightarrow SL=0$