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Step-by-Step Solution
Step 1: Understanding the Beat Frequency
When two sound sources with frequencies $f_1$ and $f_2$ are played together, the resultant sound waves interfere, producing beats. The beat frequency $f_b$ is given by:
$ f_b = |\,f_1 - f_2\,| $
Here, the tuning fork’s frequency is $512\,\text{Hz}$ and the piano string’s frequency is $f_\text{piano}$. The initial beat frequency observed is $4\,\text{Hz}$.
Step 2: Determining Possible Frequencies of the Piano String
Since the beat frequency is $4\,\text{Hz}$, the piano string’s frequency must be either:
$ f_\text{piano} = 512 \pm 4 $
Therefore, the two possible values are:
$ f_\text{piano} = 512 + 4 = 516\,\text{Hz} $
$ f_\text{piano} = 512 - 4 = 508\,\text{Hz} $
Step 3: Effect of Increasing Tension on Frequency
Increasing the tension in a piano string raises its frequency. After the tension is increased, the beat frequency decreases from $4\,\text{Hz}$ to $2\,\text{Hz}$. This means the difference between the tuning fork’s frequency ($512\,\text{Hz}$) and the new piano frequency becomes smaller:
$ f_\text{new} - 512 = 2 \quad \text{or} \quad 512 - f_\text{new} = 2 $
Thus, the new piano string frequency after tightening is closer to $512\,\text{Hz}$ than before:
If the original piano frequency had been $516\,\text{Hz}$, increasing tension would make it even higher (e.g., $517$ or $518\,\text{Hz}$, etc.), which would increase the beat frequency above $4\,\text{Hz}$, contradicting the question.
If the original frequency was $508\,\text{Hz}$, increasing tension raises it toward $512\,\text{Hz}$. This reduces the beat frequency to $2\,\text{Hz}$, matching the given scenario.
Step 4: Conclusion
Because an increase in tension should bring the piano frequency closer to the tuning fork’s frequency (thereby reducing the beat frequency from $4\,\text{Hz}$ to $2\,\text{Hz}$), the only consistent initial frequency of the piano string is:
$508\,\text{Hz}$
