Principle of Superposition, Interference and Beats: Practice Problem & Solution
Each of the two strings of length $51.6 \text{ cm}$ and $49.1 \text{ cm}$ are tensioned separately by $20 \text{ N}$ force. Mass per unit length of both the strings is same and equal to $1 \text{ g/m}$. When both the strings vibrate simultaneously the number of beats is: (2009)
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:
Velocity $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{20}{10^{-3}}} = 141.42 \text{ m/s}$. The frequencies are $f_1 = \frac{v}{2L_1} = \frac{141.42}{2 \times 0.516} \approx 137 \text{ Hz}$ and $f_2 = \frac{141.42}{2 \times 0.491} \approx 144 \text{ Hz}$. Number of beats = $144 - 137 = 7$.
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