Question

: A microphone of mass cross-section area $0.80\,{\text{c}}{{\text{m}}^2}$ is placed in front of a small speaker emitting $3.0\,{\text{W}}$ of sound output. If the distance between the microphone and the speaker is $2.0\,{\text{m}}$ , how much energy falls on the microphone in $5.0\,{\text{s}}$ ?

Answer 1

First of all, we will convert the unit of area into S.I unit and find the formula of power of sound emitted by the speaker. Then we will find the intensity of sound followed by the power of sound falling on the microphone. Lastly, we will manipulate and find energy.

Complete step by step solution:
In the given question, we are supplied the following data:
There is a microphone whose area of cross section is $0.80\,{\text{c}}{{\text{m}}^2}$ and is placed in front of a small speaker.Power of sound output by the speaker is $3.0\,{\text{W}}$.The distance between the microphone and the speaker is $2.0\,{\text{m}}$.We are asked to find the energy which falls on the microphone in $5.0\,{\text{s}}$.To begin with, we will first convert the unit of area into S.I units. We also need to calculate the intensity of sound along with power.The power of emitted by the speaker is given by the formula, which is shown below:
${P_0} = 10\pi \,{\text{W}}$
Since, the cross-sectional area $\left( a \right)$ is given as $0.80\,{\text{c}}{{\text{m}}^2}$, we will convert it into S.I units:
$0.80\,{\text{c}}{{\text{m}}^2} \\\ \Rightarrow 0.80 \times 1\,{\text{cm}} \times 1\,{\text{cm}} \\\ \Rightarrow 0.80 \times {10^{ - 2}}\,{\text{m}} \times {10^{ - 2}}\,{\text{m}} \\\ \Rightarrow 0.80 \times {10^{ - 4}}\,{{\text{m}}^2}$
We know that the sound emitted by the speaker is in the hemispherical shape, so the area of the hemisphere is written as:
$A = 2\pi {r^2}$
We can now calculate the intensity of the sound at a distance of $2.0\,{\text{m}}$ from the speaker with the help of the formula given below:
$I = \dfrac{{{P_0}}}{A}$ …… (1)
Where,
$I$ indicates the intensity of the sound.
${P_0}$ indicates the power of sound emitted by the speaker.
$A$ indicates the area of the hemisphere whose radius is the distance between the microphone and the speaker.
We will modify the equation (1) and we get:
$I = \dfrac{{{P_0}}}{A} \\\ \Rightarrow I = \dfrac{{{P_0}}}{{2\pi {r^2}}} \\\ \Rightarrow I = \dfrac{{10\pi }}{{2\pi \times {2^2}}} \\\ \Rightarrow I = \dfrac{{10}}{8}$
Now, we calculate the power of the sound which falls on the microphone, with the formula:
${P_f} = I \times a$ …… (2)
Now, substitute the required values in the equation (2), and we get:
${P_f} = I \times a \\\ \Rightarrow {P_f} = \dfrac{{10}}{8} \times 0.80 \times {10^{ - 4}} \\\ \Rightarrow {P_f} = 1 \times {10^{ - 4}}\,{\text{W}}$
Therefore, the power of the sound which falls on the microphone is $1 \times {10^{ - 4}}\,{\text{W}}$.Now, we calculate the energy of the sound which falls on the speaker in $5.0\,{\text{s}}$ with the help of the formula:
$E = {P_f} \times t$ …… (3)
Substituting the required values in the equation (3) and we get:
$E = {P_f} \times t \\\ \Rightarrow E = 1 \times {10^{ - 4}} \times 0.5 \\\ \therefore E = 0.5 \times {10^{ - 4}}\,{\text{J}}$

Hence, the energy which falls on the microphone in $5.0\,{\text{s}}$ is $0.5 \times {10^{ - 4}}\,{\text{J}}$.

Note: While solving the problem, remember that the sound emitted from the speaker is in hemispherical shape. Use of circular shape will definitely affect the result and produce irrelevant results. Conversion of units to the S.I system is must.