A tuning forke is in unison with a resonance column of length 17 cm. When the length is increased by 1 mm, three beats are heard in one second. What is the frequency of the fork? Neglect the end correction?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between frequency and length
We know that the frequency (f) of a tuning fork is inversely proportional to the length (L) of the resonance column when tension (T) and mass per unit length (μ) are constant. This can be expressed as:
\[ f \propto \frac{1}{L} \]
This means that:
\[ f_1 L_1 = f_2 L_2 \]
where \( f_1 \) and \( L_1 \) are the frequency and length of the first condition, and \( f_2 \) and \( L_2 \) are the frequency and length of the second condition.

### Step 2: Set up the known values
From the problem:
- Initial length \( L_1 = 17 \) cm = 0.17 m
- Increased length \( L_2 = 17.1 \) cm = 0.171 m (since 1 mm = 0.001 m)
- Let the frequency of the tuning fork be \( f \) (in Hz).
- The frequency corresponding to the increased length will be \( f - 3 \) Hz (since 3 beats are heard).

### Step 3: Apply the frequency-length relationship
Using the relationship established in Step 1:
\[ f \cdot L_1 = (f - 3) \cdot L_2 \]
Substituting the values:
\[ f \cdot 0.17 = (f - 3) \cdot 0.171 \]

### Step 4: Expand and rearrange the equation
Expanding the equation gives:
\[ 0.17f = 0.171f - 0.513 \]
Now, rearranging this yields:
\[ 0.171f - 0.17f = 0.513 \]
\[ 0.001f = 0.513 \]

### Step 5: Solve for the frequency
Now, divide both sides by 0.001:
\[ f = \frac{0.513}{0.001} = 513 \text{ Hz} \]

### Conclusion
The frequency of the tuning fork is **513 Hz**.

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