An accurate and reliable audio oscillator is used to standardize a tuning fork. When the oscillator reading is `514Hz`, two beats are hear per second. When the oscillator reading is `510Hz`, the beat frequency is `6Hz`. The frequency of the tuning fork is

Options

A. `506`
B. `510`
C. `516`
D. `158`

Explanation

To find the frequency of the tuning fork, we can use the information provided about the beat frequencies when the oscillator is set to different frequencies.

### Step-by-Step Solution:

1. **Understanding Beat Frequency**:
The beat frequency is the absolute difference between the frequencies of two sound sources. If we denote the frequency of the tuning fork as \( f \), then the beat frequency can be expressed as:
\[ \text{Beat Frequency} = |f - f_{oscillator}| \]

2. **First Scenario**:
When the oscillator is set to \( 514 \, \text{Hz} \), we hear \( 2 \, \text{beats/second} \):
\[ |f - 514| = 2 \]
This gives us two possible equations:
\[ f - 514 = 2 \quad \text{or} \quad f - 514 = -2 \]
Solving these:
- From \( f - 514 = 2 \):
\[ f = 516 \, \text{Hz} \]
- From \( f - 514 = -2 \):
\[ f = 512 \, \text{Hz} \]

3. **Second Scenario**:
When the oscillator is set to \( 510 \, \text{Hz} \), we hear \( 6 \, \text{beats/second} \):
\[ |f - 510| = 6 \]
This gives us two more equations:
\[ f - 510 = 6 \quad \text{or} \quad f - 510 = -6 \]
Solving these:
- From \( f - 510 = 6 \):
\[ f = 516 \, \text{Hz} \]
- From \( f - 510 = -6 \):
\[ f = 504 \, \text{Hz} \]

4. **Finding the Correct Frequency**:
Now we have two possible frequencies from the first scenario: \( 516 \, \text{Hz} \) and \( 512 \, \text{Hz} \), and two possible frequencies from the second scenario: \( 516 \, \text{Hz} \) and \( 504 \, \text{Hz} \).

The only frequency that appears in both scenarios is \( 516 \, \text{Hz} \).

5. **Conclusion**:
Therefore, the frequency of the tuning fork is:
\[ f = 516 \, \text{Hz} \]