Question #37681

1 Answer
Dec 20, 2014

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.