Question

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.

Answer 1

Correct Answer: 6

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:

\(f_1 = 6 \, \text{Hz}\)

Answer 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:

\[ f_1 = f_{\text{fork}} \]

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.