# Question 37681

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) $\setminus \mu$ is held at a tension of $T$, waves in the string propagate at a speed given by,
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad v = \setminus \sqrt{\frac{T}{\setminus} \mu}$
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.

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad {f}_{b e a t} = | {f}_{t e s t} - {f}_{r e f} |$

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 $\left(1\right)$ : ${f}_{1} / {f}_{2} = {L}_{2} / {L}_{1} = \frac{122 c m}{120 c m} = \frac{61}{60} - \left(2\right)$

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;#

$\setminus q \quad \setminus q \quad \setminus q \quad {f}_{1} / {f}_{2} = \frac{{f}_{o} + 2}{{f}_{o} - 2} - \left(3\right)$

Comparing Equations $\left(2\right)$ & $\left(3\right)$, $\setminus q \quad \frac{{f}_{o} + 2}{{f}_{o} - 2} = \frac{60}{61}$.

Solving this equation for ${f}_{o}$ we get ${f}_{o} = 242$ Hz.