Question

Question #37681

Answer 1

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.