Question #37681

1 Answer

Answer:

242 Hz.

Explanation:

Solution : The frequency of the first wire must be #242# Hz.

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 #v# is a constant, the frequency of a particular mode (#f_n#) depends only on the string length (#L#). A shorter string produces a higher frequency while a long string produces a lower frequency.

#f_n(L_1) = n/L_1.v/2; \qquad f_n(L_2) = n/L_2.v/2; \qquad (f_n(L_1))/(f_n(L_2))=L_2/L_1 - (1)#

  • 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.

#\qquad \qquad \qquad \qquad \qquad \qquad f_{beat}=|f_{test}-f_{ref}|#

This Problem : Let #f_o# be the frequency of the reference string, which is what we are interested in finding.

As the length of the test string increases from #L_1=120# cm to #L_2=122# cm, its frequency decreases from #f_1# to #f_2#.

From Equation #(1)# : #f_1/f_2=L_2/L_1 = (122 cm)/(120 cm) = 61/60 - (2)#

Since there are two beats produced in both cases, the reference frequency #f_o# must be #2# Hz lower than #f_1# and #2# Hz higher than #f_2#.#\qquad \qquad f_1 = f_{o}+2; \qquad f_2=f_o-2;#

# \qquad \qquad \qquad f_1/f_2=(f_o + 2)/(f_o - 2) - (3)#

Comparing Equations #(2)# & #(3)#, #\qquad (f_o + 2)/(f_o - 2) = 60/61#.

Solving this equation for #f_o# we get #f_o = 242# Hz.