Question

Two tuning forks produce 6 beats per second when sounded together. One of the fork is in unison with 1.5 m length of wire and the other with 2.0 m length of wire. The frequencies of forks are:

• A. 18 Hz and 22 Hz
• B. 16 Hz and 22 Hz
• C. 24 Hz and 18 Hz
• D. 6 Hz and 12 Hz

Hint:

For sonometer problems, the key relationship to remember is \( f \propto 1/L \). This means a shorter wire produces a higher pitch (higher frequency), and a longer wire produces a lower pitch. This helps in correctly setting up the beat frequency equation (\(f_{higher} - f_{lower}\)).

Answer 1

The Correct Option is C

Step 1: Conceptual Foundation: This problem integrates two physics principles: the phenomenon of beats arising from the superposition of two sound waves with slightly differing frequencies, and the inverse relationship between the fundamental frequency of a vibrating string (sonometer wire) and its length.
Step 2: Governing Equations and Methodology:
1. Beat Frequency Formula: \( f_{\text{beat}} = |f_1 - f_2| \), where \( f_1 \) and \( f_2 \) represent the frequencies of the two sound sources.
2. Sonometer Frequency Relation: \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \). Under constant tension (\(T\)) and uniform linear mass density (\(\mu\)), the frequency (\(f\)) is inversely proportional to the length (\(L\)). This relationship is expressed as:
\[ f \propto \frac{1}{L} \implies \frac{f_1}{f_2} = \frac{L_2}{L_1} \]
Step 3: Detailed Analysis:
Let \( f_1 \) and \( f_2 \) be the frequencies of the two tuning forks.
The given beat frequency is 6 Hz:
\[ |f_1 - f_2| = 6 \, \text{Hz} \]
Assume \( f_1 \) is the frequency of the fork resonating with the wire of length \( L_1 = 1.5 \) m.
Assume \( f_2 \) is the frequency of the fork resonating with the wire of length \( L_2 = 2.0 \) m.
Due to the inverse proportionality between frequency and length, the shorter wire corresponds to the higher frequency. Therefore, \( f_1>f_2 \).
The beat frequency equation can thus be written as:
\[ f_1 - f_2 = 6 \quad \text{(Equation 1)} \]
Using the sonometer relationship:
\[ \frac{f_1}{f_2} = \frac{L_2}{L_1} = \frac{2.0 \, \text{m}}{1.5 \, \text{m}} = \frac{4}{3} \]
This yields:
\[ f_1 = \frac{4}{3} f_2 \quad \text{(Equation 2)} \]
Solving the system of equations by substituting Equation 2 into Equation 1:
\[ \frac{4}{3} f_2 - f_2 = 6 \]
\[ \left(\frac{4}{3} - 1\right) f_2 = 6 \]
\[ \frac{1}{3} f_2 = 6 \]
\[ f_2 = 18 \, \text{Hz} \]
Calculating \( f_1 \) using Equation 1:
\[ f_1 = f_2 + 6 = 18 + 6 = 24 \, \text{Hz} \]
Step 4: Conclusion:
The frequencies of the two tuning forks are 24 Hz and 18 Hz. This matches option (C).