A tuning fork of frequency $512 ext{ Hz}$ makes $4 ext{ beats/s}$ with the vibrating strings of a piano. T

Question

A tuning fork of frequency $512 	ext{ Hz}$ makes $4 	ext{ beats/s}$ with the vibrating strings of a piano. The beat frequency decreases to $2 	ext{ beats/s}$ when the tension in the piano strings is slightly increased. The frequency of the piano string before increasing the tension was:

Accepted Answer

1. **Determine possible initial frequencies:** The frequency of the tuning fork is $f = 512 	ext{ Hz}$. The beat frequency is $4 	ext{ beats/s}$. Therefore, the initial frequency of the piano string ($f_p$) can be either $512 + 4 = 516 	ext{ Hz}$ or $512 - 4 = 508 	ext{ Hz}$ .
2. **Analyze the effect of tension:** When the tension in a string increases, its fundamental frequency increases because frequency is directly proportional to the square root of tension ($f \propto \sqrt{T}$) .
3. **Determine the correct frequency:**
   - If the initial frequency was $516 	ext{ Hz}$, increasing the tension would increase the frequency further (e.g., to $518 	ext{ Hz}$). The new beat frequency with the $512 	ext{ Hz}$ tuning fork would be $|518 - 512| = 6 	ext{ beats/s}$, which represents an increase in the beat frequency.
   - If the initial frequency was $508 	ext{ Hz}$, increasing the tension would increase the frequency closer to $512 	ext{ Hz}$ (e.g., to $510 	ext{ Hz}$). The new beat frequency would be $|510 - 512| = 2 	ext{ beats/s}$, which is a decrease and matches the condition given in the problem.
Therefore, the original frequency of the piano string was $508 	ext{ Hz}$.

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