Principle of Superposition, Interference and Beats: Practice Problem & Solution
A source of unknown frequency gives $4 \text{ beats/s}$, when sounded with a source of known frequency $250 \text{ Hz}$. The second harmonic of the source of unknown frequency gives five beats per second, when sounded with a source of frequency $513 \text{ Hz}$. The unknown frequency is: (2013)
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:
Unknown frequency $f = 250 \pm 4 = 254 \text{ Hz}$ or $246 \text{ Hz}$. For the second harmonic, $2f$ must give 5 beats with 513 Hz. If $f = 254$, $2f = 508 \implies |513 - 508| = 5$. If $f = 246$, $2f = 492 \implies |513 - 492| = 21$. So $f = 254 \text{ Hz}$.
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