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en/3.4.19.md
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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] | ||
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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] | |||