Solution

Statement

$3.4.19.$ The graph of coordinate versus time for a motion that is the sum of two harmonic oscillations is shown in the figure. Use it to determine the amplitudes and frequencies of these oscillations.

Solution

1. Extracting Amplitudes from the Graphical Envelope
The given graph illustrates a classic beating phenomenon resulting from the linear superposition of two harmonic oscillations with nearly equal amplitudes and slightly different frequencies ($\omega_1\approx\omega_2$).
The general equation describing the displacement $x(t)$ of such a combined system is:
$$x(t)=a_1\sin(\omega_1t)+a_2\sin(\omega_2t)$$

Looking closely at the graph parameters:
— Maximum Envelope Peak ($A$): This occurs when the two individual oscillations interfere constructively:
$$A_{\max}=a_1+a_2=A$$
— Minimum Envelope Neck ($B$): This occurs when the two components interfere destructively, opposing each other:
$$A_{\min}=|a_1-a_2|=B$$

We can solve this simple algebraic system to isolate the distinct amplitudes $a_1$ and $a_2$ (assuming $a_1>a_2$):
$$a_1=\frac{A+B}{2}, \quad a_2=\frac{A-B}{2}$$

2. Extracting Frequencies from the Graphical Periods
The graph defines two specific time intervals on the horizontal axis:
— $\tau$ (Period of Fast Oscillations): This corresponds to the time it takes to complete one full cycle of the rapid inner wave. It is directly tied to the average carrier frequency:
$$\tau=\frac{2\pi}{\omega_{\text{avg}}}=\frac{4\pi}{\omega_1+\omega_2}$$

— $T$ (Beat Period): This represents the time interval between two consecutive minimums (necks) of the slow amplitude envelope. The beat frequency is related to the difference between the two frequencies:
$$T=\frac{2\pi}{\omega_{\text{beat}}}=\frac{2\pi}{|\omega_1-\omega_2|}$$

3. Mathematical Combination to Isolate $\omega_1$ and $\omega_2$
From our period relations, we can construct expressions for the sum and difference of the target frequencies:
$$\omega_1+\omega_2=\frac{4\pi}{\tau}$$
$$\omega_1-\omega_2=\frac{2\pi}{T}$$

To isolate $\omega_1$, we add the two equations and divide by 2:
$$2\omega_1=\frac{4\pi}{\tau}+\frac{2\pi}{T} \implies \omega_1=\pi\left(\frac{2}{\tau}+\frac{1}{T}\right)$$

To isolate $\omega_2$, we subtract the second equation from the first and divide by 2:
$$2\omega_2=\frac{4\pi}{\tau}-\frac{2\pi}{T} \implies \omega_2=\pi\left(\frac{2}{\tau}-\frac{1}{T}\right)$$

Converting to cyclical frequency $f=\frac{\omega}{2\pi}$ yields:
$$f_1=\frac{1}{\tau}+\frac{1}{2T}, \quad f_2=\frac{1}{\tau}-\frac{1}{2T}$$