Solution for Question 11
Sound reflection can be modeled efficiently using an “image source” technique.
The biker honks at point $A$, which is at a distance $l$ from the wall. The sound reflects off the wall and reaches the biker at point $B$ after time $t$.
This is geometrically equivalent to sound traveling directly from the image of A (let’s call it $A’$) to the point $B$.
$A’$ is located at distance $l$ “behind” the wall. The line $AA’$ is perpendicular to the wall.
Consider the triangle formed by $A$, $A’$, and $B$. We can define the sides as follows:
- Side $AA’ = 2l$
- Side $AB = vt$ (Distance traveled by biker)
- Side $A’B = ct$ (Distance traveled by sound)
We need the angle $\angle A’AB$. Since the biker moves at an angle $\theta$ with the wall, and the line $AA’$ is perpendicular to the wall, the angle between the biker’s path $AB$ and the line $AA’$ is: $$ \alpha = 90^\circ + \theta $$
In $\triangle AA’B$, we apply the Law of Cosines on angle $\angle A’AB$: $$ (A’B)^2 = (AA’)^2 + (AB)^2 – 2(AA’)(AB) \cos(90^\circ + \theta) $$ Substituting the physical values into the equation: $$ (ct)^2 = (2l)^2 + (vt)^2 – 2(2l)(vt) (-\sin \theta) $$ $$ c^2 t^2 = 4l^2 + v^2 t^2 + 4lvt \sin \theta $$
Rearrange the equation into a standard quadratic form to solve for $t$: $$ (c^2 – v^2)t^2 – (4lv \sin \theta)t – 4l^2 = 0 $$ Using the quadratic formula: $$ t = \frac{4lv \sin \theta \pm \sqrt{(4lv \sin \theta)^2 – 4(c^2 – v^2)(-4l^2)}}{2(c^2 – v^2)} $$ Simplifying the discriminant (term under the root): $$ \Delta = 16l^2 v^2 \sin^2 \theta + 16l^2(c^2 – v^2) $$ $$ \Delta = 16l^2 (v^2 \sin^2 \theta + c^2 – v^2) = 16l^2 (c^2 – v^2 (1 – \sin^2 \theta)) $$ $$ \Delta = 16l^2 (c^2 – v^2 \cos^2 \theta) $$ Taking the root: $$ \sqrt{\Delta} = 4l \sqrt{c^2 – v^2 \cos^2 \theta} $$