Question

A flute player hears four beats per second when she compares her note to a 784 Hz tuning fork (the note G). She can match the frequency of the tuning fork by pulling out the "tuning joint" to lengthen her flute slightly. What was her initial frequency?

Answer 1

Answer:

her initial frequency is 788 Hz

Explanation:

Given;

beat frequency, Fb = 4 Hz

frequency of turning fork, F1 = 784 Hz

frequency of the flute, or her initial frequency, F2 = ?

Beat frequency is always equal to the difference in frequency of the two notes that interfere to produce the beats. In this case, to determine the frequency of the flute, which is her initial frequency, we apply this formula for beat frequency;

Fb = F2 - F1

4 = F2 - 784

F2 = 4 + 784

f2 = 788 Hz

Therefore, her initial frequency is 788 Hz