Principle of Superposition, Interference and Beats: Practice Problem & Solution
A source of sound gives 5 beats per second, when sounded with another source of frequency $100 \text{ second}^{-1}$. The second harmonic of the source, together with a source of frequency $205 \text{ sec}^{-1}$ gives 5 beats per second. What is the frequency of the source? (1995)
Solution Explained:
To solve this problem, we apply the core principles of Principle of Superposition, Interference and Beats. Understanding the underlying formula is key to arriving at the correct answer below:
Frequency $f = 100 \pm 5 = 105$ or $95 \text{ Hz}$. Second harmonic $2f = 210$ or $190 \text{ Hz}$. Since $|2f - 205| = 5$, $2f$ must be $210$. Thus $$f = 105 \text{ second}^{-1}$$.
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