Question

You have two small, identical boxes that generate 440 Hz

notes. While holding one, you drop the other from a 20-m-high

balcony. How many beats will you hear before the falling box hits

the ground? You can ignore air resistance.

Short answer

The solution gives a detailed description of air resistance and

two small, identical boxes that generate 440 Hz

notes

Step-by-step solution

Description of resistance

Resistance is a measure of opposite current flow in an electrical circuits

Step 2; Description on the frequency, time and period

Provided solutions

The integral beat of frequency

$N={\int }_0^f{f}_B\left(t\right)tdt$

The heard frequency will be$f-\frac{f_0}{1+{\displaystyle \frac{v_x}{v}}}=f_0\left(\frac{v}{v+v_s}\right)$

as per the strategic box the frequency difference is ;

$\begin{array}{rcl}& & f_{b=f_o-f_o}-f_{o\frac{v}{v+v_s}=(1-\frac{v}{v+v_s})f_o}\end{array}$

The integral part becomes

$$\begin{gathered} \frac{du}{dt}=g\stackrel{}{\to dt=\frac{du}{g}} \\ \int \frac{du}{u}=\ln \left|u\right|+C \\ N=440.(\frac{2^2}{2}-\frac{343}{9.8}.2+\frac{{343}^2}{9.8^2}\ln \left|\frac{343+9.8.2}{343}\right| \end{gathered}$$