Question #37681
1 Answer
242 Hz.
Explanation:
Solution : The frequency of the first wire must be
Explanation : Solving this problem requires knowing the following concepts,

Speed of propagation of a wave in a string : When a string of linear mass density (mass per unit length)
#\mu# is held at a tension of#T# , waves in the string propagate at a speed given by,
#\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad v=\sqrt{ T/\mu}#
Since the tension (#T# ) of the wire is held constant the speed of propagation of the wave along the string (#v# ) is also a constant. 
Standing Waves & Harmonics : The wavelength and frequencies of the various modes of standing waves (harmonics) are related to the string length (
#L# ) as:
#\qquad \lambda_n = (2L)/n; \qquad f_n=v/\lambda_n=n(v)/(2L)=(n)/(L).(v)/(2)#
Because
 Formation of Beats : Beats form as a result of the superposition of two waves that differ slightly in their frequencies. The beat frequency is equal to the frequency difference between the interacting waves.
This Problem : Let
As the length of the test string increases from
From Equation
Since there are two beats produced in both cases, the reference frequency
Comparing Equations
Solving this equation for