Principle of Superposition, Interference and Beats: Practice Problem & Solution
In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency $6 \text{ Hz}$. When tension in B is slightly decreased, the beat frequency increases to $7 \text{ Hz}$. If the frequency of A is $530 \text{ Hz}$, the original frequency of B will be : (2020)
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 of string B is $f_B = f_A \pm 6 = 530 \pm 6$. It can be $536 \text{ Hz}$ or $524 \text{ Hz}$. Decreasing tension lowers frequency. If $f_B = 524$, lowering it increases the difference from 530, producing a higher beat frequency ($7 \text{ Hz}$). Thus, original $f_B$ was $524 \text{ Hz}$.
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