A loudspeaker that produces signals from `50` to `500 Hz` is placed at the open end of a closed tube of length `1.1 m`. The lowest and the highest frequency that excites resonance in the tube are `f_(1) and f_(h)` respectively . The velocity of sound is `330 m//s`. Then

Options

A. (a)`f_(1) = 50 Hz`
B. (b)`f_(h) = 500 Hz`
C. (c)`f_(1) = 75 Hz`
D. (d)`f_(h) = 450 Hz`

Explanation

To solve the problem, we need to find the lowest frequency \( f_1 \) and the highest frequency \( f_h \) that can excite resonance in a closed tube of length \( L = 1.1 \, \text{m} \) when the speed of sound \( v = 330 \, \text{m/s} \).

### Step-by-Step Solution

1. **Understanding Resonance in a Closed Tube**:
In a closed tube, resonance occurs at odd harmonics. The frequency of the \( n \)-th harmonic is given by the formula:
\[ f_n = \frac{n \cdot v}{4L} \]
where \( n \) can take values \( 1, 3, 5, \ldots \).

2. **Calculate the Lowest Frequency \( f_1 \)**:
For the first harmonic (\( n = 1 \)):
\[ f_1 = \frac{1 \cdot 330}{4 \cdot 1.1} \]
\[ f_1 = \frac{330}{4.4} = 75 \, \text{Hz} \]

3. **Calculate the Highest Frequency \( f_h \)**:
The highest frequency \( f_h \) that can be produced by the loudspeaker is \( 500 \, \text{Hz} \). We need to find the maximum \( n \) such that \( f_n \leq 500 \, \text{Hz} \):
\[ 500 = \frac{n \cdot 330}{4 \cdot 1.1} \]
Rearranging gives:
\[ n = \frac{500 \cdot 4 \cdot 1.1}{330} \]
\[ n = \frac{2200}{330} \approx 6.67 \]
Since \( n \) must be an odd integer, we take \( n = 6 \) (the largest odd integer less than 6.67).

4. **Calculate \( f_h \) for \( n = 6 \)**:
\[ f_h = \frac{6 \cdot 330}{4 \cdot 1.1} \]
\[ f_h = \frac{1980}{4.4} = 450 \, \text{Hz} \]

### Final Answers
- The lowest frequency \( f_1 = 75 \, \text{Hz} \)
- The highest frequency \( f_h = 450 \, \text{Hz} \)