Question

Loudness of sound coming from a point source at a distance R is 50 dB. What will be the new loudness value if the distance is doubled?
(A) 25 dB
(B) 12.5 dB
(C) 100 dB
(D) 44 dB

Answer 1

Hint: Convert the intensity of sound to power per unit area and then find the new intensity at 2R distance which can be converted back to decibels. The intensity of 0 dB sound is 10$^{ - 12}$W/m$^2$. This intensity is denoted as I$_0$

Complete step-by-step solution
The loudness of sound at distance R is given in decibels. We represent this loudness as I$_{LR}$. In order to find out the loudness of sound at 2R distance, we first need to find the power of the source. Power is intensity of sound multiplied by area.
The intensity of the sound at distance R can be denoted as I$_R$. Using the formula for loudness of sound, we obtain,
${I_{LR}} = 10\log \dfrac{{{I_R}}}{{{I_0}}}$ = 50 dB
We substitute the value of I$_0$ to find the value of I$_R$, we obtain,
${I_R} = {I_0}anti\log \left( {\dfrac{{50}}{{10}}} \right) = {10^{ - 12}}anti\log \left( 5 \right)$ …equation (1)
On substituting the antilog(5) = 10$^5$ in equation (1), we get,
${I_R} = {10^{ - 12 + 5}} = {10^{ - 7}}$ W/m$^2$
Now, we find the power of the source. To find the power of the source, we multiply the intensity with area = 4$\pi {R^2}$ which is equal to the surface area of a sphere of radius R since the source is a point which emits power in all directions. The intensity of sound at a particular distance from the sound will be the same in all directions.
Power of point source, P = 10$^{ - 7}$$ \times 4\pi {R^2}$ W
Now, we find the intensity of sound at a distance of 2R by dividing the power of point source by the area of sphere of radius 2R. Let us denote the intensity of sound at distance 2R as I$_{2R}$.
${I_{2R}} = \dfrac{{{{10}^{ - 7}} \times 4\pi {R^2}}}{{4\pi {{(2R)}^2}}}$ W/m$^2$
We simplify this expression to obtain the value of the intensity of sound at 2R distance.
${I_{2R}} = 0.25 \times {10^{ - 7}}W/{m^2}$
The loudness of sound at 2R distance can be found by using the formula for loudness of sound. Let us denote the loudness of sound at 2R distance as I$_{L\left( {2R} \right)}$. We obtain,
${I_{L\left( {2R} \right)}} = 10\log \left( {\dfrac{{{I_{2R}}}}{{{I_0}}}} \right)$
Substituting the values, we obtain,
${I_{L\left( {2R} \right)}} = 10\log \left( {\dfrac{{0.25 \times {{10}^{ - 7}}}}{{{{10}^{ - 12}}}}} \right) = 10\log \left( {0.25 \times {{10}^5}} \right)$
Using the property of logarithm to the base 10, we can write the right side of the expression as the sum of logarithms of the product of the two terms. We also know that log of 10 to the power of any number is the power of 10.
${I_{L(2R)}}$ = 10[log(0.25) + 5] = 10(-0.602 + 5) = 10 $ \times $ 4.398 dB
Hence, we obtain that the loudness of sound at distance 2R is, ${I_{L(2R)}}$ = 43.98 dB $ \approx $44 dB

Therefore, the correct option is D.

Note: The power from a source radiates in radial direction. Hence, the intensity changes radially. This concept can be used in many other questions as well.