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

Respuesta :

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:
[tex]f_b = |f_1 - f_2|[/tex]
where [tex]f_b[/tex] is the beat frequency, and [tex]f_1, f_2[/tex] are the frequencies of the two forks.

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