Question

A small speaker delivers 2W of audio output. At what distance from the speaker will one detect 120dB intensity sound? [Given reference intensity of sound as 10−12W/m2].
(A) 10 cm
(B) 30 cm
(C) 40 cm
(D) 20 cm

Answer 1

Hint:-This given problem can be solved by taking the consideration of loudness of sound that is coming from any source such as the speaker.

Complete step-by-step solution:
Step 1:
As we know that loudness of sound in dB (decibels) is given by the formula –
\[dB = 10\log \dfrac{I}{{\mathop I\nolimits_0 }}\] (1)
Where \[I = \]Given intensity of sound from the speaker, and
\[\mathop I\nolimits_0 = \]Reference intensity of sound i.e., given as \[\mathop I\nolimits_0 = \mathop {10}\nolimits^{ - 12} \]W/m2
And it is also given in the question that the dB intensity of sound is 120 dB at some distance from the speaker.
So, by keeping all these values in equation (1), value of the intensity of sound from the speaker can be calculated –
\[120 = 10\log \dfrac{I}{{\mathop {10}\nolimits^{ - 12} }}\] on further solving this
\[12 = \log \dfrac{I}{{\mathop {10}\nolimits^{ - 12} }}\]
Now, the base of this logarithm is 10-base so to remove logs we have to raise both sides to the same exponent as the base of the logarithm.
So, \[\mathop {10}\nolimits^{12} = \mathop {10}\nolimits^{\log \left( {\dfrac{I}{{\mathop {10}\nolimits^{ - 12} }}} \right)} \] on further solving this
\[\mathop {10}\nolimits^{12} = \dfrac{I}{{\mathop {10}\nolimits^{ - 12} }}\]
\[I = \mathop {10}\nolimits^{12} \times \mathop {10}\nolimits^{ - 12} \]
\[I = 1\]W/m2.


Step 2: But we know that sound intensity, also known as acoustic intensity, can be defined as the power carried by the sound waves per unit area and these sound waves will be perpendicular in direction to that area.
So, \[I = \dfrac{P}{A}\] (2)
The SI unit of intensity is the watt per square meter (W/m2).
So, for the above question it is given that speaker delivers 2W of audio output power (power radiated) and it is also known that the power radiated from the speaker (sound energy from the source) is in a spherical form and hence area A is equal to the area of sphere with radius R.
And, area of sphere is (say), \[A = 4\pi \mathop R\nolimits^2 \] (3)
So this denotes the intensity of sound at a distance R from the speaker.
From equation (2) and (3), we will get –
\[I = \dfrac{P}{{4\pi \mathop R\nolimits^2 }}\] on rearranging this equation
\[\mathop R\nolimits^2 = \dfrac{P}{{4\pi I}}\] on substituting the above values in the equation
\[\mathop R\nolimits^2 = \dfrac{2}{{4 \times 3.14 \times 1}}\]
\[\mathop R\nolimits^2 = 0.1592\]m2
\[R = \sqrt {0.1592} \]m
\[R = 0.3989 \simeq 0.399\]m
\[R = 40\]m.
So, the distance at which intensity of sound will be 120dB is 40cm.
So, the correct option is (C).

Note:- Loudness is referred to as how loud or soft a sound seems to a listener. The loudness of sound is determined by the intensity of the sound waves. Intensity is a measure of the amount of energy in sound waves.