A sound source sends a sinusoidal sound wave of angular frequency $\mathbf{\text{3000rad/s}}$ and amplitude $\mathbf{\text{12.0m}}$through a tube of air. The internal radius of the tube is$\mathbf{\text{2.00cm}}$ .

(a) What is the average rate at which energy (the sum of the kinetic and potential energies) is transported to the opposite end of the tube?

(b) If, simultaneously, an identical wave travels along an adjacent, identical tube, what is the total average rate at which energy is transported to the opposite ends of the two tubes by the waves? If, instead, those two waves are sent along the same tube simultaneously, what is the total average rate at which they transport energy when their phase difference is

(c) $\mathbf{0}$

(d)$\mathbf{\text{0.40πrad}}$ ,

(e)$\mathbf{\text{π\hspace{0.17em}rad}}$?

• Amplitude,$\mathrm{A}=12\text{\hspace{0.17em}nm}$
• Radius,$\mathrm{r}=12\text{\hspace{0.17em}cm}$
• Density of air,$\mathrm{\rho }=1.21\frac{\text{kg}}{{\text{m}}^{\text{3}}}$
• Angular frequency,$\mathrm{\omega }=3000\frac{\text{rad}}{\text{s}}$
• Speed of sound, $\mathrm{v}=343\text{\hspace{0.17em}m/s}$

Superposition principle is applied to get the resultant amplitude. Use the formula for power for the wave to find the power at the given instant.

Formulae are as follow:

1. Power $\mathrm{P}=\frac{1}{2}{\mathrm{\rho A}}^2{\mathrm{v\omega }}^2\left({\mathrm{\pi r}}^2\right)$
2. ${\mathrm{A}}^{\text{'}}=\mathrm{A}\sqrt{2(1+\mathrm{cos\phi })}$

Where, $P$ is power, $A$ is area, $\rho$ is density, $v$ is velocity, $\omega$ is angular frequency and $r$ is radius.

Average rate of energy transfer is defined as,

$\mathrm{P}=\frac{1}{2}{\mathrm{\rho A}}^2{\mathrm{v\omega }}^2\left(\text{π}{\mathrm{r}}^2\right)$

$\begin{array}{c}\mathrm{P}=0.5\left(\frac{\text{1.21kg}}{{\text{m}}^{\text{3}}}\right){(12\times {10}^{-9})}^2\left(\frac{\text{343m}}{\text{s}}\right){\left(\frac{\text{3000rad}}{\text{s}}\right)}^2{\left(3.14)\times (2\times {10}^{-2}\text{m}\right)}^2\\ =0.34\times {10}^{-9}\text{W}\\ =0.34\text{\hspace{0.17em}nW}\end{array}$

Hence, the average rate of energy transferred is $P=0.\text{34nW}$.

Since, the waves and tube are identical, so, net area will double. So, energy transfer from the other side will be double its initial,

$\begin{array}{c}\mathrm{P}=2\left(0.3\text{4nW}\right)\\ =0.68\text{nW}\end{array}$

Hence, the energy transferred by two pipes, $\mathrm{P}=0.68\text{nW}$.

Resultant amplitude is defined as,

${\mathrm{A}}^{\text{'}}=\mathrm{A}\sqrt{2(1+\mathrm{cos\phi })}$

For $\mathrm{\phi }=0,$

$\begin{array}{c}{\mathrm{A}}^{\text{'}}=\mathrm{A}\sqrt{2(1+\cos 0^0)}\\ =\mathrm{A}\sqrt{4}\\ =2\mathrm{A}\end{array}$

$\mathrm{P}=\frac{1}{2}\mathrm{\rho }{{\mathrm{A}}^{\text{'}}}^2{\mathrm{v\omega }}^2\left({\mathrm{\pi r}}^2\right)$

$\mathrm{P}=4\left(\frac{1}{2}{\mathrm{\rho A}}^2{\mathrm{v\omega }}^2\left({\mathrm{\pi r}}^2\right)\right)$

$\begin{array}{c}\mathrm{P}=\text{4}\left(\text{0.34nW}\right)\\ =1.3\text{6nW}\\ \approx 1.4\text{nW}\end{array}$

Hence, the energy transferred when$\phi =0$ is $P=1.4\text{nW}$.

$\begin{array}{c}{\mathrm{A}}^{\text{'}}=\mathrm{A}\sqrt{2(1+\cos 0.4\mathrm{\pi })}\\ =1.62\mathrm{A}\end{array}$

$\begin{array}{c}\mathrm{P}=\frac{1}{2}\mathrm{\rho }{{\mathrm{A}}^{\text{'}}}^2{\mathrm{v\omega }}^2\left({\mathrm{\pi r}}^2\right)\\ =\frac{1}{2}\mathrm{\rho }{\left(1.62\right)}^2{\mathrm{A}}^2{\mathrm{v\omega }}^2\left({\mathrm{\pi r}}^2\right)\\ =2.63\left(\frac{1}{2}{\mathrm{\rho A}}^2{\mathrm{v\omega }}^2\left({\mathrm{\pi r}}^2\right)\right)\\ =2.63\left(0.\text{34nW}\right)\\ =0.8\text{8nW}\end{array}$

Hence, the energy transferred when $\phi =0.\text{4π}$ is $P=0.88\text{nW}$.

$\begin{array}{c}{\mathrm{A}}^{\text{'}}=\mathrm{A}\sqrt{2(1+\mathrm{cos\pi }}\\ =\mathrm{A}\sqrt{2(1-1)}\\ =0\end{array}$

Then,

$\text{Power}=\mathrm{P}=0$

Hence,the energy transferred when$\mathrm{\phi }=\text{πrad}$, is zero.

Therefore, the superposition principle can be used to find the resultant amplitude and then apply the formula for power.