31. Tuning fork 1 (420 Hz) is sounded with tuning fork 2 and 20 beats are count in 10s.

Question

a) What are the possible frequencies of tuning fork 2?
b) Tuning fork 2 is sounded with tuning fork 3 (426 Hz) and 12 beats are detected in 3s. what are the possible frequencies of tuning fork 2?
c) Using both (a) and (b), what is the frequency of turning fork 2? Explain your reasoning.

Answer 1

a) To find the possible frequencies of tuning fork 2, we can use the formula:

f2 = (f1 + Δf) or (f1 - Δf)

Where f2 is the frequency of tuning fork 2, f1 is the frequency of tuning fork 1, and Δf is the difference in frequency between the two forks.

Given that the frequency of tuning fork 1 (f1) is 420 Hz and 20 beats are counted in 10 seconds, we can calculate Δf:

Δf = (beat count / time) * 2

Δf = (20 / 10) * 2 = 4 Hz

Substituting the values into the formula, we have:

f2 = 420 + 4 = 424 Hz (possible frequency)
f2 = 420 - 4 = 416 Hz (possible frequency)

b) To find the possible frequencies of tuning fork 3, we use the same formula:

f3 = (f2 + Δf) or (f2 - Δf)

Given that the frequency of tuning fork 2 (f2) is 424 Hz and 12 beats are detected in 3 seconds, we can calculate Δf:

Δf = (beat count / time) * 2

Δf = (12 / 3) * 2 = 8 Hz

Substituting the values into the formula, we have:

f3 = 424 + 8 = 432 Hz (possible frequency)
f3 = 424 - 8 = 416 Hz (possible frequency)

c) To determine the frequency of tuning fork 2, we need to find the value of f2 that is common to both parts a) and b).
From part a), we found two possible frequencies for tuning fork 2: 424 Hz and 416 Hz.
From part b), we found two possible frequencies for tuning fork 2: 432 Hz and 416 Hz.

Comparing the values, we see that the only frequency that appears in both parts a) and b) is 416 Hz. Therefore, the frequency of tuning fork 2 is 416 Hz.