When two tuning forks are stuck simultaneously, 3 beats per second are heard. The frequency of one fork is 510 Hz. A piece of wax is placed on the 510 Hz fork to lower its frequency slightly. If the beat frequency is increased, what is
the correct frequency for the second fork?
Answer in units of Hz
To find the correct frequency for the second fork, we need to understand how beats are produced. Beats occur when two sound waves with slightly different frequencies interfere with each other. The beat frequency is equal to the difference in frequencies between the two tuning forks.
Given that when the two tuning forks are struck simultaneously, 3 beats per second are heard, we can calculate the difference in frequencies. Let's call the frequency of the second fork "f" (in Hz).
f - 510 Hz = 3 beats per second
To solve for the frequency "f", we rearrange the equation:
f = 3 beats per second + 510 Hz
f = 3 Hz + 510 Hz
f = 513 Hz
Therefore, the correct frequency for the second fork is 513 Hz.
From the original beat frequency of 3 Hz, you know that the "other" fork has a frequency of either 507 or 513 Hz.
If you lower the frequency of the 510 Hz fork and get a higher beat frequency, the "other" fork must have the higher 513 Hz frequency, NOT 507.