Final answer:
The frequency of the unknown tuning fork before clay was added is 438 Hz as this value generates a beat frequency of 2 Hz initially and after decreasing by 1 Hz, the beat frequency changes to 3 Hz, which is consistent with the problem statement.
Step-by-step explanation:
The student is asking about the frequency of a tuning fork given the beat frequency when sounded with a tuning fork of known frequency. Beat frequency arises when two sounds of slightly different frequencies interfere with each other, which is a concept from physics. Initially, the unknown tuning fork has a beat frequency of 2 Hz with a fork of 440 Hz. After adding clay and decreasing the frequency of the unknown fork by 1 Hz, the beat frequency becomes 3 Hz.
To find the frequency of the other tuning fork, we need to consider two possibilities due to the nature of beats: the unknown frequency could be higher or lower than 440 Hz. Therefore, we have either 440 Hz + beat frequency or 440 Hz - beat frequency.
In the first case, the other tuning fork could have been 440 Hz + 2 Hz = 442 Hz, but after placing the clay, the frequency would decrease to 441 Hz, thus making the new beat frequency 440 Hz - 441 Hz = -1 Hz (which is not a possibility since beat frequency is always a positive number).
In the second case, the fork could have been 440 Hz - 2 Hz = 438 Hz originally, and after placing the clay, it would become 437 Hz, which would yield a beat frequency of 440 Hz - 437 Hz = 3 Hz. This matches the situation described after placing the clay.
Therefore, the correct answer is C. 438 Hz, which was the frequency of the second tuning fork before clay was added.