Principle of Superposition, Interference and Beats: Practice Problem & Solution
A source of frequency $v$ gives 5 beats/second when sounded with a source of frequency $200 \text{ Hz}$. The second harmonic of frequency $2v$ of source gives 10 beats/second when sounded with a source of frequency $420 \text{ Hz}$. The value of $v$ is: (1994)
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 $v = 200 \pm 5 = 205$ or $195 \text{ Hz}$. Also $|2v - 420| = 10$. If $v = 205$, then $2v = 410$, and $|410 - 420| = 10$, which matches. Thus $v = 205 \text{ Hz}$.
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