91Ó°ÊÓ

• Amplitude,$\mathrm{A}=12\text{ n³¾}$
• Radius,$\mathrm{r}=12\text{ c³¾}$
• Density of air,$\mathrm{ÒÏ}=1.21\frac{\text{kg}}{{\text{m}}^{\text{3}}}$
• Angular frequency,$\mathrm{Ó\neg }=3000\frac{\text{rad}}{\text{s}}$
• Speed of sound, $\mathrm{v}=343\text{″¾/²õ}$

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 $\mathrm{P}=\frac{1}{2}{\mathrm{ÒÏA}}^2{\mathrm{vÓ\neg }}^2\left({\mathrm{Ï€°ù}}^2\right)$
2. ${\mathrm{A}}^{\text{'}}=\mathrm{A}\sqrt{2(1+\mathrm{³¦´Ç²õφ})}$

Where, $P$ is power, $A$ is area, $\mathrm{ÒÏ}$ is density, $v$ is velocity, $\mathrm{Ó\neg }$ is angular frequency and $r$ is radius.

Average rate of energy transfer is defined as,

$\mathrm{P}=\frac{1}{2}{\mathrm{ÒÏA}}^2{\mathrm{vÓ\neg }}^2\left(\text{Ï€}{\mathrm{r}}^2\right)$

$\begin{array}{c}\mathrm{P}=0.5\left(\frac{\text{1.21kg}}{{\text{m}}^{\text{3}}}\right){(12×{10}^{−9})}^2\left(\frac{\text{343m}}{\text{s}}\right){\left(\frac{\text{3000rad}}{\text{s}}\right)}^2{\left(3.14)×(2×{10}^{−2}\text{m}\right)}^2\\ =0.34×{10}^{−9}\text{W}\\ =0.34\text{ n°Â}\end{array}$

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,

${\mathrm{A}}^{\text{'}}=\mathrm{A}\sqrt{2(1+\mathrm{³¦´Ç²õφ})}$

For $\mathrm{φ}=0,$

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

$\mathrm{P}=\frac{1}{2}\mathrm{ÒÏ}{{\mathrm{A}}^{\text{'}}}^2{\mathrm{vÓ\neg }}^2\left({\mathrm{Ï€°ù}}^2\right)$

$\mathrm{P}=4\left(\frac{1}{2}{\mathrm{ÒÏA}}^2{\mathrm{vÓ\neg }}^2\left({\mathrm{Ï€°ù}}^2\right)\right)$

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

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

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

$\begin{array}{c}\mathrm{P}=\frac{1}{2}\mathrm{ÒÏ}{{\mathrm{A}}^{\text{'}}}^2{\mathrm{vÓ\neg }}^2\left({\mathrm{Ï€°ù}}^2\right)\\ =\frac{1}{2}\mathrm{ÒÏ}{\left(1.62\right)}^2{\mathrm{A}}^2{\mathrm{vÓ\neg }}^2\left({\mathrm{Ï€°ù}}^2\right)\\ =2.63\left(\frac{1}{2}{\mathrm{ÒÏA}}^2{\mathrm{vÓ\neg }}^2\left({\mathrm{Ï€°ù}}^2\right)\right)\\ =2.63\left(0.\text{34nW}\right)\\ =0.8\text{8nW}\end{array}$

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

$\begin{array}{c}{\mathrm{A}}^{\text{'}}=\mathrm{A}\sqrt{2(1+\mathrm{³¦´Ç²õÏ€}}\\ =\mathrm{A}\sqrt{2(1−1)}\\ =0\end{array}$

Then,

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

Hence,the energy transferred when$\mathrm{φ}=\text{Ï€°ù²¹»å}$, is zero.

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