Amplitude of Beats

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Theorem

Let $W_1$ and $W_2$ be harmonic waves whose frequencies are $f_1$ and $f_2$.

Let the amplitude of $W_1$ and $W_2$ both be $a$.

Let the superpositon of $W_1$ onto $W_2$ exhibit the phenomenon of beats.

The amplitude $A_b$ of those beats at time $t$ is:

$A_b = 2 a \map \cos {\pi \size {f_1 - f_2} t - \dfrac \epsilon 2}$


Proof

Let $\omega_1$ and $\omega_2$ denote the angular frequency of $W_1$ and $W_2$ respectively.

Let us consider the harmonic waves that are $W_1$ and $W_2$ as they disturb the medium at $x = 0$.

Without loss of generality, therefore, let $W_1$ and $W_2$ be be expressed as:

\(\ds \map {\phi_1} t\) \(=\) \(\ds a \sin \omega_1 t\)
\(\ds \map {\phi_2} t\) \(=\) \(\ds a \map \sin {\omega_2 t + \epsilon}\)

where:

\(\ds \omega_1\) \(=\) \(\ds 2 \pi f_1\) Definition of Angular Frequency
\(\ds \omega_2\) \(=\) \(\ds 2 \pi f_2\)

and $\epsilon$ is the phase of $\phi_1$ with respect to $\phi_2$.


Then:

\(\ds \map {\phi_1} t + \map {\phi_2} t\) \(=\) \(\ds a \sin \omega_1 t + a \map \sin {\omega_2 t + \epsilon}\) superimposing the two waves
\(\ds \) \(=\) \(\ds 2 a \map \sin {\dfrac {\omega_1 t + \omega_2 t + \epsilon} 2} \map \cos {\dfrac {\omega_1 t - \paren {\omega_2 t + \epsilon} } 2}\) Sine plus Sine
\(\ds \) \(=\) \(\ds 2 a \map \sin {\dfrac {\paren {\omega_1 + \omega_2} t + \epsilon} 2} \map \cos {\dfrac {\paren {\omega_1 - \omega_2} t - \epsilon} 2}\) simplifying
\(\ds \) \(=\) \(\ds 2 a \map \sin {\dfrac {\paren {\omega_1 + \omega_2} t + \epsilon} 2} \map \cos {\dfrac {\size {\omega_1 - \omega_2} t - \epsilon} 2}\) Cosine Function is Even
\(\ds \leadsto \ \ \) \(\ds \size {\map {\phi_1} t + \map {\phi_2} t}\) \(\le\) \(\ds \size {2 a \map \cos {\dfrac {\size {\omega_1 - \omega_2} t - \epsilon} 2} }\) as $\map \sin {\dfrac {\paren {\omega_1 + \omega_2} t} 2} \le 1$ throughout


Hence the amplitude $a$ of the slower component is seen to be:

$A_b = \dfrac {\size {\omega_1 - \omega_2} } 2$

By definition of angular frequency:

\(\ds A_b\) \(=\) \(\ds 2 a \map \cos {\dfrac {\size {\omega_1 - \omega_2} t - \epsilon} 2}\)
\(\ds \) \(=\) \(\ds 2 a \map \cos {\dfrac {\size {2 \pi f_1 - 2 \pi f_2} t - \epsilon} 2}\) Definition of Angular Frequency
\(\ds \) \(=\) \(\ds 2 a \map \cos {\pi \size {f_1 - f_2} t - \dfrac \epsilon 2}\)

Hence the result.

$\blacksquare$


Sources

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