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Q104P

Expert-verifiedFound in: Page 513

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Suppose a spherical loudspeaker emits sound isotropically at ${\mathbf{10}}{\mathbf{}}{\mathit{W}}$** ** into a room with completely absorbent walls, floor, and ceiling (an anechoic chamber). (a) What is the intensity of the sound at distance ${\mathit{d}}{\mathbf{=}}{\mathbf{}}{\mathbf{3}}{\mathbf{.0}}{\mathbf{}}{\mathit{m}}$**

- The intensity of the sound at distance ${d}_{2}=3.0\text{m}$ from the center of the source is $0.088\text{\hspace{0.17em}}\frac{\text{W}}{{\text{m}}^{\text{2}}}$
- The ratio of the wave amplitude ${d}_{1}=4.0\text{m}$ to that at ${d}_{2}=3.0\text{m}$ is $0.75$

- The power of the spherical loudspeaker is ${P}_{s}=10\text{\hspace{0.17em}W}$
- The point of intensity from source of sound is ${d}_{1}=4.0\text{\hspace{0.17em}m}$
- The point of intensity from source of sound is ${d}_{2}=3.0\text{\hspace{0.17em}m}$

**Use the concept of variation of intensity with distance. The intensity of the sound from an isotropic point source decreases with the square of the distance from the source. The intensity is proportional to the square of the amplitude.**

Formulae:

The intensity of the wave, $I=\frac{{P}_{s}}{4\pi {r}^{2}}$ (i)

The intensity of a wave is directly proportional to the square of the amplitude,

$I\propto {A}^{2}$ (ii)

Using equation (i), the intensity of the sound wave at 3.0 m is given as:

$\begin{array}{c}I=\frac{10W}{4\times 3.14\times {(3.0\text{m})}^{2}}\\ =0.088\frac{\text{W}}{{\text{m}}^{\text{2}}}\end{array}$

Hence, the value of the intensity of the sound is $0.088\frac{\text{W}}{{\text{m}}^{\text{2}}}$

The intensity of sound is proportional to the square of the amplitude. The intensity of sound at distance ${d}_{1}=4.0m$ is ${I}_{1}$ and wave amplitude is ${A}_{1}$ . Hence, using equation (ii), the intensity relation is given as:

${I}_{1}\propto {A}_{1}^{2}$, ${I}_{2}\propto {A}_{2}^{2}$ ….. (1)

The intensity of sound at distance ${d}_{2}=3.0m$ is ${I}_{2}$ and wave amplitude is ${A}_{2}$ . Hence, using equation (ii), the intensity relation is given as

${I}_{2}\propto {A}_{2}^{2}$ ….. (2)

Divide equation (1) by equation (2), we get

$\begin{array}{c}\frac{{I}_{1}}{{I}_{2}}=\frac{{A}_{1}^{2}}{{A}_{2}^{2}}\\ \frac{{A}_{1}}{{A}_{2}}=\sqrt{\frac{{I}_{1}}{{I}_{2}}}\end{array}$ ….. (3)

Using equation (i), the equation (3) becomes:

$\begin{array}{c}\frac{{A}_{1}}{{A}_{2}}=\sqrt{\frac{\left(\frac{{P}_{s}}{4\pi {r}_{1}^{2}}\right)}{\left(\frac{{P}_{s}}{4\pi {r}_{2}^{2}}\right)}}\\ =\frac{{r}_{2}}{{r}_{1}}\\ =\frac{3.0m}{4.0m}\\ =0.75\end{array}$

Hence, the value of the ratio of amplitudes is $0.75$.

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