A tube `1.0 m` long is closed at one end. A stretched wire is placed near the open end. The wire is `0.3 m` long and a mass of `0.01 kg` . It is held fixed at both ends and vibrates in its fundamental mode. It sets the air column in the tube into vibration at its fundamental frequency by resonance. Find
(a) the frequency of oscillation of the air column and
(b) the tension in the wire.
Speed of sound in air = `330 m//s` .

Explanation

To solve the problem step by step, we will first find the frequency of oscillation of the air column and then calculate the tension in the wire.

### Step 1: Understand the relationship between the frequencies of the air column and the wire
In this problem, we have a tube closed at one end (the air column) and a stretched wire vibrating in its fundamental mode. The frequencies of both the air column and the wire will be the same due to resonance.

### Step 2: Calculate the frequency of the air column
The fundamental frequency \( F_2 \) of the air column can be calculated using the formula:
\[ F_2 = \frac{V_2}{4L_1} \]
where:
- \( V_2 \) is the speed of sound in air (given as \( 330 \, \text{m/s} \)),
- \( L_1 \) is the length of the air column (which is \( 1.0 \, \text{m} \)).

Substituting the values:
\[ F_2 = \frac{330}{4 \times 1.0} = \frac{330}{4} = 82.5 \, \text{Hz} \]

### Step 3: Calculate the frequency of the wire
The fundamental frequency \( F_1 \) of the wire can be calculated using the formula:
\[ F_1 = \frac{V_1}{2L_2} \]
where:
- \( V_1 \) is the speed of the wave on the wire,
- \( L_2 \) is the length of the wire (which is \( 0.3 \, \text{m} \)).

Since the frequencies are equal (\( F_1 = F_2 \)), we can set up the equation:
\[ F_1 = F_2 = 82.5 \, \text{Hz} \]

### Step 4: Calculate the speed of the wave on the wire
From the frequency formula for the wire, we can express the speed \( V_1 \):
\[ V_1 = 2L_2 \cdot F_1 \]
Substituting the known values:
\[ V_1 = 2 \times 0.3 \times 82.5 = 49.5 \, \text{m/s} \]

### Step 5: Calculate the tension in the wire
The speed of the wave on the wire is related to the tension \( T \) and the linear mass density \( \mu \) of the wire:
\[ V_1 = \sqrt{\frac{T}{\mu}} \]
Rearranging gives:
\[ T = \mu V_1^2 \]
First, we need to calculate the linear mass density \( \mu \):
\[ \mu = \frac{m}{L_2} = \frac{0.01 \, \text{kg}}{0.3 \, \text{m}} = \frac{0.01}{0.3} \approx 0.0333 \, \text{kg/m} \]

Now substituting \( V_1 \) and \( \mu \) into the tension formula:
\[ T = 0.0333 \times (49.5)^2 \approx 0.0333 \times 2450.25 \approx 81.67 \, \text{N} \]

### Final Results
(a) The frequency of oscillation of the air column is \( 82.5 \, \text{Hz} \).

(b) The tension in the wire is approximately \( 81.67 \, \text{N} \).