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yKpoI14uk [10]
3 years ago
11

Four beats per second are heard when an unknown tuning fork and a 240 Hz tuning fork are struck simultaneously. This same unknow

n tuning fork and a 250 Hz tuning fork are struck. Six beats are heard. Its exact frequency: is
256 Hz is
236 Hz is
244 Hz
cannot be determined
Physics
1 answer:
Scilla [17]3 years ago
5 0
When the tuning forks are struck simultaneously, the beat frequency is equal to the absolute value of the difference between the frequencies of the two tuning forks:
f_b = |f_1 - f_2|
where f_b is the beat frequency, and f_1, f_2 are the frequencies of the two forks.

In the first part of the problem, f_1 = 240 Hz and f_b=4 Hz, with f_2 being the frequency of the unknown fork. In the second part of the problem, f_1=250 Hz and f_2 = 6 Hz. So we have the following system of two equations:
4 Hz = |240 Hz -f_2|
6 Hz = |250 Hz - f_2|
and the solution of this system is f_2 = 244 Hz
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