If the nearest object in front of the detector is too far away, the echo will not get back before a second click is emitted. Once that happens, the computer has no way of knowing that the echo isn't an echo from the second click and the detector doesn't give correct results anymore. Once the speaker emits a click, how much time does the echo have to return to the microphone before the next click is emitted

t = 2x / v    ( time echo),       t = 2.9 10⁻² s

Explanation:

In this case we can use the uniform motion relationships, since the sound wave has a constant speed. Let's start by calculating the time it takes for the click to reach the detector.

          v = d / t

           t = d / v

where d is the distance from the speaker to the detector and v the speed of sound (v = 340 m / s)

Now let's analyze the echo, it is produced by a reflection of the sound from a large obstacle in the direction of the sound, therefore if the distance to the obstacle is x, the echo travels a distance of 2x in this time (to)

            2x = v to

            2x = v (d / v)

            d = 2x

if we substitute in the first equation

            t = 2x / v    ( time echo)

Let's analyze these results, if the distance relationship is fulfilled, the detector (microphone) is not able to distinguish between a click and the echo of the previous click

For a numerical result suppose that the distance from the loudspeaker to the detector is d = 10 m, we obtain that the obstacle must be at a distance from the loudspeaker of

                x = 5 m

                t = 2  5/ 340

                t = 2.9 10⁻² s

            This is the time the echo has to return in this speaker-microphone configuration

because if the tool that you are using is not balanced it there would be false mass.

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach