Two tuning forks when sounded together produce 3 beats per second. On loading one of them with a little wax, 20 beats are heard in 4 seconds. Find its frequency if that of other is 386.

Answer

To solve the problem, we need to determine the frequency of one tuning fork when we know the frequency of the other and the beat frequencies produced under different conditions.

### Step-by-Step Solution:

1. **Understanding Beats**:
When two tuning forks are sounded together, the number of beats per second (beat frequency) is equal to the absolute difference between their frequencies. If we denote the frequency of the first tuning fork as \( f_1 \) and the frequency of the second tuning fork as \( f_2 \), the beat frequency \( f_b \) can be expressed as:
\[ f_b = |f_1 - f_2| \]

2. **Given Information**:
- The beat frequency when both forks are sounded together is 3 beats per second.
- The frequency of the second tuning fork (let's say \( f_2 \)) is given as 386 Hz.
- When one fork is loaded with wax, 20 beats are heard in 4 seconds, which gives a new beat frequency of:
\[ f_b' = \frac{20 \text{ beats}}{4 \text{ seconds}} = 5 \text{ beats per second} \]

3. **Setting Up Equations**:
From the first condition (3 beats per second):
\[ |f_1 - 386| = 3 \]
This gives us two possible equations:
\[ f_1 - 386 = 3 \quad \text{or} \quad 386 - f_1 = 3 \]
Thus, we can solve for \( f_1 \):
- Case 1: \( f_1 = 386 + 3 = 389 \) Hz
- Case 2: \( f_1 = 386 - 3 = 383 \) Hz

4. **Considering the Loaded Fork**:
When one fork is loaded with wax, the new beat frequency is 5 beats per second. This can be expressed as:
\[ |f_1' - 386| = 5 \]
Here \( f_1' \) is the new frequency of the first fork after loading with wax. The possible equations are:
\[ f_1' - 386 = 5 \quad \text{or} \quad 386 - f_1' = 5 \]
Solving these gives:
- Case 1: \( f_1' = 386 + 5 = 391 \) Hz
- Case 2: \( f_1' = 386 - 5 = 381 \) Hz

5. **Finding the Correct Frequency**:
We need to determine which of the two original frequencies \( f_1 \) (389 Hz or 383 Hz) corresponds to the new frequencies after loading with wax (391 Hz or 381 Hz):
- If \( f_1 = 389 \) Hz, then the new frequency \( f_1' \) would be:
\[ f_1' = 389 - \text{(some decrease due to wax)} \]
This does not lead to either 391 Hz or 381 Hz.
- If \( f_1 = 383 \) Hz, then the new frequency \( f_1' \) would be:
\[ f_1' = 383 + \text{(some increase due to wax)} \]
This can lead to 381 Hz, which is plausible.

6. **Conclusion**:
Therefore, the frequency of the first tuning fork \( f_1 \) is:
\[ \boxed{383 \text{ Hz}} \]