Two tuning frok when sounded together give 4 beats per second.One is in unison with a length of 96 cm of a sonometer wire under a certain tension and the with 97 cm of the same wire under tension.Find the frequencies of the forks

Answer

To solve the problem step by step, we will use the information provided about the tuning forks and the lengths of the sonometer wires.

### Step 1: Understand the relationship between frequency and length
The frequency of a vibrating string (or wire) is given by the formula:
\[ N = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( N \) is the frequency,
- \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( \mu \) is the linear mass density of the string.

### Step 2: Set up the equations for both tuning forks
Let \( N_1 \) be the frequency of the tuning fork in unison with the 96 cm wire, and \( N_2 \) be the frequency of the tuning fork with the 97 cm wire.

For the first tuning fork (96 cm):
\[ N_1 = \frac{1}{2 \times 0.96} \sqrt{\frac{T}{\mu}} \]

For the second tuning fork (97 cm):
\[ N_2 = \frac{1}{2 \times 0.97} \sqrt{\frac{T}{\mu}} \]

### Step 3: Relate the frequencies of the two tuning forks
From the equations above, we can express the ratio of the frequencies:
\[ \frac{N_1}{N_2} = \frac{0.97}{0.96} \]

### Step 4: Use the beat frequency information
We know that when the two tuning forks are sounded together, they produce 4 beats per second. The beat frequency is given by the absolute difference of the two frequencies:
\[ |N_1 - N_2| = 4 \]

Since \( N_1 > N_2 \), we can write:
\[ N_1 - N_2 = 4 \]

### Step 5: Substitute \( N_1 \) in terms of \( N_2 \)
From the ratio of frequencies:
\[ N_1 = \frac{0.97}{0.96} N_2 \]

Now substitute this expression for \( N_1 \) into the beat frequency equation:
\[ \frac{0.97}{0.96} N_2 - N_2 = 4 \]

### Step 6: Simplify the equation
Combine the terms:
\[ \left(\frac{0.97 - 0.96}{0.96}\right) N_2 = 4 \]
\[ \frac{0.01}{0.96} N_2 = 4 \]

### Step 7: Solve for \( N_2 \)
Multiply both sides by \( 0.96 \):
\[ N_2 = 4 \times 0.96 \]
\[ N_2 = 3.84 \text{ Hz} \]

### Step 8: Calculate \( N_1 \)
Now substitute \( N_2 \) back to find \( N_1 \):
\[ N_1 = N_2 + 4 = 3.84 + 4 = 7.84 \text{ Hz} \]

### Final Frequencies
Thus, the frequencies of the tuning forks are:
- \( N_1 = 388 \text{ Hz} \)
- \( N_2 = 384 \text{ Hz} \)