Two tuning forks `A` and `B` produce `6` beats per second. Frequency of `A` is `300 H_(Z)` . When `B` is slightly loaded with wax, beat frequency decreases . Find original frequency of `B` .

Explanation

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

### Step 1: Understand the Beat Frequency
The beat frequency is given as 6 beats per second. This means that the difference in frequencies between the two tuning forks (A and B) is 6 Hz.

### Step 2: Write the Frequency Relationship
Let the frequency of tuning fork A be \( \nu_A = 300 \, \text{Hz} \) and the frequency of tuning fork B be \( \nu_B \). The relationship for the beat frequency can be expressed as:
\[
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