Question

here is a procedure used in a middle school science class to find the speed of sound in air.
a student takes two erasers and begins to bang then together with a regular beat - perhaps at a rate of 20 claps in 10 second - in front of the brick wall of the school building. the student begins walking backward until she cant hear the echo of the erasers anymore.
explain, in terms of the principles of waves explored in this lesson, how this procedure is able to estimate the speed of sound in air.

Explanation:

When the student claps the erasers, sound waves are produced. These waves travel to the brick wall and reflect back as an echo. The student walks backward until the echo coincides with the next clap. At this point, the time it takes for the sound to travel to the wall and back (the round - trip time) is equal to the time between claps. The rate of claps (e.g., 20 claps in 10 seconds) gives the time interval between claps. If we assume the distance from the student to the wall when the echo is no longer distinguishable is \(d\), and the time between claps is \(t\), the speed of sound \(v\) can be calculated using the formula \(v=\frac{2d}{t}\) (the factor of 2 is because the sound travels to the wall and back).

Answer:

The procedure works by using the principle that when the echo aligns with the next clap, the time for the sound's round - trip (to the wall and back) equals the clap interval. Using the formula \(v = \frac{2d}{t}\) (where \(d\) is the distance from the student to the wall at the critical point and \(t\) is the clap interval), the speed of sound \(v\) can be estimated.