Principle of Superposition, Interference and Beats: Practice Problem & Solution
Two sound waves with wavelengths $5.0 \text{ m}$ and $5.5 \text{ m}$ respectively, each propagate in a gas with velocity $330 \text{ m/s}$. We expect the following number of beats per second: (2006)
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:
The frequencies are $f_1 = \frac{v}{\lambda_1} = \frac{330}{5.0} = 66 \text{ Hz}$ and $f_2 = \frac{v}{\lambda_2} = \frac{330}{5.5} = 60 \text{ Hz}$. Number of beats = $f_1 - f_2 = 66 - 60 = 6$.
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