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Step-by-Step Solution
Step 1: Understand the concept of beats
Beats occur when two waves of slightly different frequencies interfere. The beat frequency is the difference in frequencies of the two interfering waves. Mathematically,
$ \text{Beat frequency} = |f_1 - f_2| $
Step 2: Identify known and unknown variables
Frequency of known tuning fork = $340 \text{ Hz}$
Frequency of unknown tuning fork = $f_A$ (to be determined)
Initial beat frequency = $5 \text{ beats/s}$
Beat frequency after filing = $2 \text{ beats/s}$
Step 3: Set up equations from initial information
When the two forks produce $5 \text{ beats/s}$ initially, it implies:
$|f_A - 340| = 5$
Thus, $f_A$ could be either:
$f_A = 340 + 5 = 345 \text{ Hz}$ or $f_A = 340 - 5 = 335 \text{ Hz}$
Step 4: Use the effect of filing on the unknown fork
Filing the tuning fork slightly increases its frequency. Let the new, slightly increased frequency of fork A be $f'_A$. After filing, the beat frequency decreases from $5 \text{ Hz}$ to $2 \text{ Hz}$. Hence:
$|f'_A - 340| = 2$
So the new frequency $f'_A$ is either:
$f'_A = 340 + 2 = 342 \text{ Hz}$ or $f'_A = 340 - 2 = 338 \text{ Hz}$
Since filing increases the fork's frequency, $f'_A > f_A$. Therefore, if $f_A$ were $345 \text{ Hz}$ to begin with, filing would make it even larger, not smaller than $345 \text{ Hz}$. But that would not match with any of the new possible frequencies ($338 \text{ Hz}$ or $342 \text{ Hz}$) that give a beat frequency of $2 \text{ Hz}$ against $340 \text{ Hz}$.
On the other hand, if $f_A$ were $335 \text{ Hz}$ initially, filing it could raise it to $338 \text{ Hz}$ or $342 \text{ Hz}$. The latter ($342 \text{ Hz}$) is a plausible new frequency, producing $2 \text{ beats/s}$ with $340 \text{ Hz}$. Thus, the initial value of $f_A$ must have been $335 \text{ Hz}$.
Step 5: State the final answer
The frequency of the unknown tuning fork $A$ is $335 \text{ Hz}$.
