Two simple harmonic progressive waves have time periods of $5 \mathrm{~s}$ and $4 \mathrm{~s}$ respe

Question

Two simple harmonic progressive waves have time periods of $5 \mathrm{~s}$ and $4 \mathrm{~s}$ respectively. If they are superimposed, the frequency of the resultant beats produced per second will be

Professor PC Thomas Classes · Answer

(a) : The frequencies of the two waves are the reciprocals of their time periods. Therefore, $f_1 = \frac{1}{T_1} = \frac{1}{5} = 0.20 \mathrm{~Hz}$ and $f_2 = \frac{1}{T_2} = \frac{1}{4} = 0.25 \mathrm{~Hz}$. The beat frequency is the difference between the frequencies of the two superimposed waves: $f_{beat} = |f_1 - f_2| = 0.25 - 0.20 = 0.05 \mathrm{~Hz}$.