A tuning fork of unknows frequency makes three beats per second with a standard fork of frequency `384 H_(Z)`. The beat frequency decreases when a small piece of wax is put on a prong of the first fork. What is the frequency of this fork?

Explanation

To find the frequency of the tuning fork of unknown frequency, we can follow these steps:

### Step 1: Understand the beat frequency
The beat frequency is the absolute difference between the frequencies of two tuning forks. In this case, we have:
- Frequency of the standard fork (f2) = 384 Hz
- Beat frequency = 3 beats per second

The relationship can be expressed as:
\[ |f_1 - f_2| = 3 \]
This means:
\[ f_1 - f_2 = 3 \quad \text{or} \quad f_2 - f_1 = 3 \]

### Step 2: Set up the equations
Since we know that the beat frequency decreases when wax is added to the first fork, it indicates that the frequency of the first fork (f1) is greater than that of the standard fork (f2). Therefore, we can write:
\[ f_1 - f_2 = 3 \]

### Step 3: Substitute the known frequency
Now, substituting the value of f2 into the equation:
\[ f_1 - 384 = 3 \]

### Step 4: Solve for f1
Now, we can solve for f1:
\[ f_1 = 3 + 384 \]
\[ f_1 = 387 \text{ Hz} \]

### Conclusion
The frequency of the tuning fork of unknown frequency is **387 Hz**.

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