Maths Quiz - Solution

If an aeroplane (e.g. the Concorde) flies with a velocity faster than the speed of sound, it flies away from the spherical sound waves it generates. When the sound waves reach the ground, the viewers on the ground get to hear an ultrasonic bang that makes windows burst. Let's say the velocity of the aeroplane, which is constantly flying at the same height, is twice as fast as the speed of sound. When the people at a certain location get to hear the ultrasonic sound, the distance from the aeroplane to the air above that location is already one kilometer. At what height is it flying?

Solution:

Of course you need some knowledge of classical physics in order to solve this task: A spherical wave can be pictured as a sphere (or, to simplify it yet more, a circle) with its radius being the wavelength. Its center point has the same distance from the ground (i.e. height) as the aeroplane. Classical physics teaches us that sound waves are growing linearly with time: l = c * t, where l ... wavelength, c ... speed of sound, t ... time.

In fact there is more than one possible solution because the task was not defined well enough. I wrote: "When the people at a certain location get to hear the ultrasonic sound...". The question here is: What ultrasonic sound?

Most people assumed that I was asking for the ultrasonic sound that was emitted when the plane was right above the watchers. In this case, the solution is as follows:

When the sound wave hits the ground, its wavelength is the distance from the center point to the ground, i.e. the height of the aeroplane. When this occurs, the aeroplane has already moved 1000 m from the air above that location. We can describe this movement of the aeroplane with a formula similar to the one of the wave: s = v * t, where s ... distance from the air above a specific location, v ... velocity of the aeroplane, t ... time.

The common variable of these two equations is t, so let's transform them: t = l / c = s / v. We need l, which is: l = s * c / v. As the aeroplane is twice as fast as sound, c / v = 1 / 2. s is 1000 m. Thus we get: l = 1000 m * 1 / 2 = 500 m.

This was rather simple. However, Boiled Brain pointed out to me that this is not the first instance of the ultrasonic sound that is audible. Yet before that, people can hear the ultrasonic sound emitted at different positions of the plane. Boiled Brain tried to find out the height of the plane if the task is about the first time the people get to hear the ultrasonic sound. Here is his solution:

"Let h be the height, s be the speed of sound, l be the distance from the air above the location to the plane (point L). Let t = 0 be the moment when l = 0. Then the plane is in the point L in the moment t = l/(2s). The distance from L to the location is d = sqrt(h^2 + l^2) and sound takes d/s seconds to pass it. So the people hear the sound made by the plane in the point L in the moment t = l/2s + d/s = (l/2 + sqrt(h^2 + l^2))/s. We are interested in the moment when they hear the sound for the first time. So we need to find the minimum of the function. Let's take its derivative:

t' = (1/2 + l/sqrt(h^2 +l^2))/s

The minimum is when t' = 0, so

l/sqrt(h^2+l^2) = -1/2 => 2l = -sqrt(h^2 + l^2) => 3l^2 = h^2 => => l = -h/sqrt(3) (l < 0)

From that moment till the moment the people hear the bang the plane goes twice as far as the sound. Hence,

1 - l = 2 sqrt(h^2 + l^2) <=> 1 + h/sqrt(3) = 2 sqrt(4/3 h^2) => => 3/sqrt(3) h = 1 => h = sqrt(3)/3

Answer: The height is sqrt(3)/3 = 0.577350269 km (approximately :))"

Congrats to Boiled Brain, mados, Stas and an anonymous CFXweb user for solving this task.