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Step-by-Step Solution
Step 1: Understanding the Concept of Beats
When two tuning forks of slightly different frequencies are struck together, they produce beats. The number of beats per second (beat frequency) is given by the absolute difference of their frequencies. Symbolically, if $n_1$ and $n_2$ are the frequencies of the two forks, the beats heard per second is:
$\Delta n = |n_1 - n_2|$
Step 2: Identify the Known Variables
Frequency of fork 1, $n_1 = 200\ \text{Hz}$
Initial number of beats heard per second, $\Delta n_{\text{initial}} = 4\ \text{Hz}$
Number of beats heard per second after attaching tape to fork 2, $\Delta n_{\text{after}} = 6\ \text{Hz}$
Unknown original frequency of fork 2, say $n_2$
Step 3: Applying Beat Frequency Before Attaching Tape
Originally, the difference in frequencies between the two forks is 4 Hz:
$|n_1 - n_2| = 4$
Since $n_1 = 200\ \text{Hz}$, $n_2$ could be either $200 + 4 = 204\ \text{Hz}$ or $200 - 4 = 196\ \text{Hz}.$
Step 4: Analyzing the Effect of Adding Tape (Loading)
When a small piece of tape is attached to a fork, it adds a slight mass, causing the frequency to generally decrease. Thus, if $n_2$ was the original frequency of fork 2, after attaching tape, its frequency becomes slightly less than $n_2.$
Step 5: Re-evaluating with the New Beat Frequency
After attaching the tape, the beat frequency increases from 4 Hz to 6 Hz. This implies that the difference between the new frequency of fork 2 and 200 Hz has become larger. Hence, if the fork’s frequency was originally below 200 Hz, loading it would lower it further, increasing the difference from 200 Hz.
Thus, instead of 204 Hz decreasing (which would bring it closer to 200 Hz and reduce the beat frequency), the fork’s original frequency must have been below 200 Hz. Therefore, the original frequency must be 196 Hz, which on further decrease upon loading results in the beat frequency rising from 4 Hz to 6 Hz.
Step 6: Conclusion
The correct original frequency of fork 2 is 196 Hz.
