A 30.0-cm-long wire having a mass of 10.0 g is fixed at the two ends and is vibrated in its fundamental mode.A 50.0-cm-long closed organ pipe, placed with its open end near the wire, is set up into resonance in its fundamental mode by the vibrating wire. Find the tension in the wire. Speed of sound in air `= 340 m s^-1`.

Explanation

To find the tension in the wire, we will follow these steps:

### Step 1: Gather the given data
- Length of the wire, \( L_w = 30.0 \, \text{cm} = 0.30 \, \text{m} \)
- Mass of the wire, \( M_w = 10.0 \, \text{g} = 0.010 \, \text{kg} \)
- Length of the closed organ pipe, \( L_p = 50.0 \, \text{cm} = 0.50 \, \text{m} \)
- Speed of sound in air, \( V = 340 \, \text{m/s} \)

### Step 2: Calculate the mass per unit length (\( \mu \)) of the wire
The mass per unit length \( \mu \) is given by:
\[ \mu = \frac{M_w}{L_w} \]
Substituting the values:
\[ \mu = \frac{0.010 \, \text{kg}}{0.30 \, \text{m}} = \frac{0.010}{0.30} = 0.0333 \, \text{kg/m} \]

### Step 3: Calculate the fundamental frequency of the wire (\( \nu_w \))
The fundamental frequency of the wire is given by:
\[ \nu_w = \frac{1}{2L_w} \sqrt{\frac{T}{\mu}} \]
We will rearrange this equation to solve for \( T \):
\[ T = \left(2L_w \nu_w\right)^2 \mu \]

### Step 4: Calculate the fundamental frequency of the closed organ pipe (\( \nu_p \))
The fundamental frequency of the closed organ pipe is given by:
\[ \nu_p = \frac{V}{4L_p} \]
Substituting the values:
\[ \nu_p = \frac{340 \, \text{m/s}}{4 \times 0.50 \, \text{m}} = \frac{340}{2} = 170 \, \text{Hz} \]

### Step 5: Set the frequencies equal (resonance condition)
Since the wire and the organ pipe are in resonance, we have:
\[ \nu_w = \nu_p \]
Thus:
\[ \frac{1}{2L_w} \sqrt{\frac{T}{\mu}} = 170 \]

### Step 6: Substitute and solve for \( T \)
Substituting \( L_w = 0.30 \, \text{m} \) and \( \mu = 0.0333 \, \text{kg/m} \):
\[ \frac{1}{2 \times 0.30} \sqrt{\frac{T}{0.0333}} = 170 \]
This simplifies to:
\[ \frac{1}{0.60} \sqrt{\frac{T}{0.0333}} = 170 \]
\[ \sqrt{\frac{T}{0.0333}} = 170 \times 0.60 = 102 \]
Squaring both sides:
\[ \frac{T}{0.0333} = 102^2 \]
\[ T = 102^2 \times 0.0333 \]
Calculating \( 102^2 \):
\[ 102^2 = 10404 \]
Thus:
\[ T = 10404 \times 0.0333 \approx 346.8 \, \text{N} \]

### Final Answer
The tension in the wire is approximately:
\[ T \approx 347 \, \text{N} \]

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