Question

A 1.0 W point source emits sound waves isotropically. Assuming that the energy of the waves is conserved, find the intensity (a) 1.0 m from the source and (b) 2.5 m from the source.

Short answer

1. The intensity 1.0 m from the source is $0.080\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$.
2. The intensity 2.5 m from the source is $0.013\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$.

Step-by-step solution

Given

1. Power,$P=1.0\text{W}$
2. Radius,$r_1=1.0\text{m}$
3. Radius,$r_2=2.5\text{\hspace{0.17em}m}$

Determining the concept

If the source emits sound wavesisotropically, it means waves are emitted in an identical direction when energy is conserved. The intensity of the sound waves can be calculated by power per unit area.

Intensity of the sound wave is given as-

$I=\frac{P}{A}$

where P is power, A is area and I is intensity.

(a) Determining the intensity of 1.0 m from the source

Since the intensity is the power emitted by the source to the area of the sphere. For radius 1.0 m²,

$\begin{array}{rcl}I& =& \frac{P}{A}\\ I& =& \frac{P}{4\pi r^2}\\ I& =& \frac{\text{1.0W}}{{\text{4×3.14×(1.0m)}}^{\text{2}}}\\ & & \begin{array}{c}I=0.0796\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\\ =0.080\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\end{array}\\ & & \\ & & \end{array}$

Hence,the intensity 1.0 m from the source is$0.080\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$ .

(b) Determining the intensity 2.5 m from the source

For radius 2.5 m,

$\begin{array}{c}I=\frac{P}{A}\\ =\frac{P}{4\pi r^2}\end{array}$

For the given values,

role="math" localid="1661492208414" $\begin{array}{l}I=\frac{\text{1.0W}}{{\text{4×3.14×(2.5m)}}^{\text{2}}}\\ =0.0127\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\\ =0.013\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\end{array}$

Hence, the intensity 2.5 m from the source is $0.013\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$.