A sonometer has `25` forks. Each produces `4` beats with the next one. If the maximum frequency is `288Hz`, which is the frequency of last fork. The lowest frequency is

Options

A. `72Hz`
B. `96Hz`
C. `128Hz`
D. `192Hz`

Explanation

To solve the problem, we need to find the lowest frequency of the last fork in a sonometer with 25 forks, given that each fork produces 4 beats with the next one, and the maximum frequency is 288 Hz.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Number of forks (n) = 25
- Number of beats (b) = 4
- Maximum frequency (f_max) = 288 Hz

2. **Understand the Relationship Between Frequencies:**
- The difference in frequency between consecutive forks can be expressed as:
\[ f_{max} - f_{min} = (n - 1) \times b \]
- Here, \(f_{min}\) is the frequency of the lowest fork.

3. **Substitute the Known Values:**
- Substitute \(n = 25\) and \(b = 4\) into the equation:
\[ f_{max} - f_{min} = (25 - 1) \times 4 \]
- This simplifies to:
\[ f_{max} - f_{min} = 24 \times 4 = 96 \]

4. **Rearranging the Equation:**
- Now we can rearrange the equation to find \(f_{min}\):
\[ f_{min} = f_{max} - 96 \]

5. **Substitute the Maximum Frequency:**
- Substitute \(f_{max} = 288\) Hz into the equation:
\[ f_{min} = 288 - 96 \]
- This simplifies to:
\[ f_{min} = 192 \text{ Hz} \]

6. **Conclusion:**
- The lowest frequency of the last fork is **192 Hz**.

### Final Answer:
The lowest frequency is **192 Hz**.

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