A certain length of wire when vibrated with another sonometer wire produced 2 beats per second for both the lengths ##122 cm## and ##120 cm## of this second sonometer wire and it's tension is same in both the occasions. What is the frequency of first wire?

242 Hz.

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 & : 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.