A tuning fork resonates with a sonometer wire of length 1 m stretched with a tension of 6 N. When the tension in the wire is changed to 54 N, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is _______ Hz.
Correct Answer: 6
Approach Solution - 1
The frequency of vibration of a sonometer wire is given by:
\(f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\)
Calculating for initial tension:
\(f_1 = \frac{1}{2} \sqrt{\frac{6}{\mu}}\)
Calculating for new tension:
\(f_2 = \frac{1}{2} \sqrt{\frac{54}{\mu}}\)
Given:
\(f_2 - f_1 = 12\)
Ratio of frequencies:
\(\frac{f_1}{f_2} = \frac{1}{3}\)
Substituting values:
\(f_1 = 6 \, \text{Hz}\)
Approach Solution -2
The problem requires finding the frequency of a tuning fork by analyzing its interaction with a sonometer wire under two different tension conditions: one of resonance and another that produces beats.
Concept Used:
1. Frequency of a Stretched String: The fundamental frequency (\( f \)) of a string vibrating in one segment is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]where \( L \) is the length of the string, \( T \) is the tension, and \( \mu \) is the linear mass density (mass per unit length).
2. Proportionality: For a given sonometer wire of fixed length (\(L\)) and linear mass density (\(\mu\)), the frequency is directly proportional to the square root of the tension:
\[ f \propto \sqrt{T} \]3. Resonance: When the tuning fork resonates with the sonometer wire, their frequencies are equal.
4. Beats: When two sound sources of slightly different frequencies are sounded together, the beat frequency (\( f_{\text{beat}} \)) is the absolute difference between their individual frequencies:
\[ f_{\text{beat}} = |f_a - f_b| \]Step-by-Step Solution:
Step 1: Define the frequencies for the initial and final conditions.
Let the frequency of the tuning fork be \( f_{\text{fork}} \).
In the initial case, the tension is \( T_1 = 6 \, \text{N} \). Let the frequency of the sonometer wire be \( f_1 \). Since the wire resonates with the tuning fork, we have:
In the final case, the tension is changed to \( T_2 = 54 \, \text{N} \). Let the new frequency of the wire be \( f_2 \). The tuning fork produces 12 beats per second with the wire, so:
\[ |f_2 - f_{\text{fork}}| = 12 \, \text{Hz} \]Step 2: Find the relationship between the two frequencies of the wire, \( f_1 \) and \( f_2 \).
Using the proportionality \( f \propto \sqrt{T} \), we can write the ratio of the frequencies:
\[ \frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}} \]Substitute the given tension values:
\[ \frac{f_2}{f_1} = \sqrt{\frac{54}{6}} = \sqrt{9} = 3 \]This gives us the relationship:
\[ f_2 = 3f_1 \]Step 3: Use the beat frequency equation to solve for the frequencies.
We have three relations: 1. \( f_1 = f_{\text{fork}} \) 2. \( |f_2 - f_{\text{fork}}| = 12 \) 3. \( f_2 = 3f_1 \)
Substitute (1) into (3):
\[ f_2 = 3f_{\text{fork}} \]Now substitute this into the beat frequency equation (2):
\[ |3f_{\text{fork}} - f_{\text{fork}}| = 12 \]Simplify the expression:
\[ |2f_{\text{fork}}| = 12 \]Final Computation & Result:
Since frequency must be a positive value, we can remove the absolute value bars.
\[ 2f_{\text{fork}} = 12 \] \[ f_{\text{fork}} = \frac{12}{2} = 6 \, \text{Hz} \]The frequency of the tuning fork is 6 Hz.
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