Question

9. A sound whispered at one focus of a whispering gallery can be heard at the other focus. The figure at the top of the next column shows a whispering gallery whose cross section is a semielliptical arch with a height of 24 feet and a width of 80 feet. How far from the room's center should two people stand so that they can whisper back and forth and be heard?

Answer 1

To whisper back and forth,

They must select two points where echo is least i.e., all the sound after bouncing from the walls reach the other person at the same time.

Now, only two points are possible for this which is foci of the ellipse

Now, if we take the center of the room as (0,0)

the equation of the full ellipse given by,

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

here, a=40ft

b=24ft

eccentricity of the ellipse

$e= \sqrt{1-\frac{b^{2}}{a^{2}}}$

$\Rightarrow e= \sqrt{1-\frac{24^{2}}{40^{2}}}$

$\Rightarrow e= \sqrt{1-\frac{9}{25}}$

$\Rightarrow e= \sqrt{\frac{16}{25}}$

$\Rightarrow e= \frac{4}{5}$

So, distance of foci from the center =$\pm ae$

$d=\pm 40*\frac{4}{5}$

$\Rightarrow d=\pm 32ft$, which is the required solution.

If any doubt feel free to comment.