91Ó°ÊÓ

• The last echo heard by someone standing just inside the door reportedly comes in 15s .
• The distance of the wall is d = 25.7 m .

The expression for the time taken to hear the echo is given by,

$\mathit{t}\mathbf{=}\frac{\mathbf{2}\mathbf{d}}{\mathbf{v}}$

Here d is the distance between the door and the person, is the speed of the sound.

The value of v is 343 m/s.

When the speed is constant,

$v=\frac{d}{t}$

Here, let v be the speed of sound. Hence,

$v=343 m/s$

Since, the factor$\frac{1}{\left(n+1\right)}$ for n = 1 reflection, t = 15 s .

Being the time for sound to travel to the far wall,

$d=vt$

Substitute$343 m/s$ for v and$\frac{\text{15}}{\text{2}}\text{s}$ for t into the above equation,

$$\begin{gathered} d=343×\frac{15}{2} \\ =2.6×1{\text{0}}^{\text{3}}  \text{m} \end{gathered}$$

Hence, the door is$2.6×1{\text{0}}^{\text{3}}  \text{m}$ far from the wall.

Similarly, the factor $\frac{1}{2}$in part was$\frac{1}{\left(n+1\right)}$ for n= 1 reflection, so, write,

d = vt

Substitute$\frac{15}{\left(n+1\right)}$ for t into the above equation,

$$\begin{gathered} d=v×\frac{15}{\left(n+1\right)} \\ \left(n+1\right)=\frac{15v}{d} \\ n=\frac{15v}{d}-1 \end{gathered}$$

Substitute 343 m/s for v and 25.7 m for d into the above equation.

$$\begin{gathered} n=\frac{15×343}{25.7}-1 \\ =199 \\ ≈2.0×{10}^2 \end{gathered}$$

Hence, the number ofreflections (back and forth) correspond to the last echo are .

$2.0×{10}^2$