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: Understanding the Concept:
This problem combines two concepts: the production of beats between two sound sources of slightly different frequencies, and the fundamental frequency of a vibrating string (sonometer wire), which is inversely proportional to its length.
Step 2: Key Formula or Approach:
1. Beat frequency: \( f_{\text{beat}} = |f_1 - f_2| \), where \( f_1 \) and \( f_2 \) are the frequencies of the two tuning forks.
2. Sonometer frequency: \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \). For a given wire under constant tension (\(T\)) and with constant linear mass density (\(\mu\)), the frequency is inversely proportional to the length (\(L\)).
\[ f \propto \frac{1}{L} \implies \frac{f_1}{f_2} = \frac{L_2}{L_1} \] Step 3: Detailed Explanation:
Let the frequencies of the two tuning forks be \( f_1 \) and \( f_2 \).
Given beat frequency is 6 Hz:
\[ |f_1 - f_2| = 6 \, \text{Hz} \] Let \( f_1 \) be the frequency of the fork in unison with the \( L_1 = 1.5 \) m wire.
Let \( f_2 \) be the frequency of the fork in unison with the \( L_2 = 2.0 \) m wire.
Since frequency is inversely proportional to length, the shorter wire corresponds to the higher frequency. Therefore, \( f_1>f_2 \).
This means we can write the beat equation as:
\[ f_1 - f_2 = 6 \quad \text{(Equation 1)} \] Now, using the sonometer relation:
\[ \frac{f_1}{f_2} = \frac{L_2}{L_1} = \frac{2.0 \, \text{m}}{1.5 \, \text{m}} = \frac{4}{3} \] \[ f_1 = \frac{4}{3} f_2 \quad \text{(Equation 2)} \] Now we solve the system of two equations. Substitute Equation 2 into Equation 1:
\[ \frac{4}{3} f_2 - f_2 = 6 \] \[ \left(\frac{4}{3} - 1\right) f_2 = 6 \] \[ \frac{1}{3} f_2 = 6 \] \[ f_2 = 18 \, \text{Hz} \] Now find \( f_1 \) using Equation 1:
\[ f_1 = f_2 + 6 = 18 + 6 = 24 \, \text{Hz} \] Step 4: Final Answer:
The frequencies of the two forks are 24 Hz and 18 Hz. This corresponds to option (C).