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+### Statement
+
+$3.4.19.$ [Insert the problem statement]
+
+### 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
+
+<b>1. Extracting Amplitudes from the Graphical Envelope</b>
+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:
+— <b>Maximum Envelope Peak ($A$):</b> This occurs when the two individual oscillations interfere constructively:
+$$A_{\max}=a_1+a_2=A$$
+— <b>Minimum Envelope Neck ($B$):</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 $a_1>a_2$):
+$$a_1=\frac{A+B}{2}, \quad a_2=\frac{A-B}{2}$$
+
+<b>2. Extracting Frequencies from the Graphical Periods</b>
+The graph defines two specific time intervals on the horizontal axis:
+— $\tau$ <b>(Period of Fast Oscillations):</b> 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:
+$$\tau=\frac{2\pi}{\omega_{\text{avg}}}=\frac{4\pi}{\omega_1+\omega_2}$$
+
+— $T$ <b>(Beat Period):</b> 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:
+$$T=\frac{2\pi}{\omega_{\text{beat}}}=\frac{2\pi}{|\omega_1-\omega_2|}$$
+
+<b>3. Mathematical Combination to Isolate $\omega_1$ and $\omega_2$</b>
+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:
+$$f_1=\frac{1}{\tau}+\frac{1}{2T}, \quad f_2=\frac{1}{\tau}-\frac{1}{2T}$$
+
+#### Answer
+$$a_1=\frac{A+B}{2}, \quad a_2=\frac{A-B}{2}$$
+$$\omega_1=\pi\left(\frac{2}{\tau}+\frac{1}{T}\right), \quad \omega_2=\pi\left(\frac{2}{\tau}-\frac{1}{T}\right)$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]