Question

Question: When the door of the Chapel of the mausoleum in Hamilton, Scotland, is slammed shut, the last echo heard by someone standing just inside the door reportedly comes 15 slater. a) If that echo were due to a single reflection off a wall opposite the door, how far from the door is the wall? (b) If, instead, the wall is 25.5 maway, how many reflection (back and forth) correspond to the last echo?

Short answer

Answer

1. The door is$2.6\times 1{\text{0}}^{\text{3}} \text{m}$far from the wall.
2. The number of reflections (back and forth) correspond to the last echo are$2.0\times {10}^2$.

Step-by-step solution

Given

• 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 .

Determining the concept

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.

(a) Determining how far from the door is the wall

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,

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

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

(b) Determining the number of reflections (back and forth) correspond to the last echo

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,

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

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

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

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

$2.0\times {10}^2$