The air in a closed tube `34 cm` long is vibrating with two nodes and two antinodes and its temprature is `51^(@)C`. What is the wavelength of the waves produced in air outside the tube, when the temperature of air is `16^(@)C`?

Options

A. `42.8 cm`
B. `68 cm`
C. `17 cm`
D. `102 cm`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the setup
We have a closed tube of length \( L = 34 \, \text{cm} \) that has two nodes and two antinodes. In a closed tube, the relationship between the length of the tube and the wavelength of the sound wave can be expressed as:
\[ L = \frac{n \lambda}{4} \]
where \( n \) is the number of quarter wavelengths fitting into the tube.

### Step 2: Determine the number of wavelengths
In this case, since there are two nodes and two antinodes, we can say that \( n = 3 \) (as there are three quarter wavelengths in the tube). Therefore, we can write:
\[ L = \frac{3 \lambda}{4} \]

### Step 3: Solve for the wavelength at 51°C
Rearranging the equation to solve for \( \lambda \):
\[ \lambda = \frac{4L}{3} \]
Substituting the value of \( L \):
\[ \lambda = \frac{4 \times 34 \, \text{cm}}{3} = \frac{136 \, \text{cm}}{3} \approx 45.33 \, \text{cm} \]

### Step 4: Convert temperatures to Kelvin
Next, we need to convert the temperatures from Celsius to Kelvin:
- For \( T = 51°C \):
\[ T = 51 + 273 = 324 \, \text{K} \]
- For \( T' = 16°C \):
\[ T' = 16 + 273 = 289 \, \text{K} \]

### Step 5: Use the relationship between wavelength and temperature
The speed of sound in air is related to temperature, and we can express the relationship between the wavelengths at different temperatures using:
\[ \frac{\lambda'}{\lambda} = \sqrt{\frac{T'}{T}} \]
where \( \lambda' \) is the wavelength at \( T' \) and \( \lambda \) is the wavelength at \( T \).

### Step 6: Substitute the known values
Substituting the known values into the equation:
\[ \frac{\lambda'}{45.33} = \sqrt{\frac{289}{324}} \]

### Step 7: Calculate \( \lambda' \)
Now, we calculate \( \sqrt{\frac{289}{324}} \):
\[ \sqrt{\frac{289}{324}} = \frac{17}{18} \]
Thus,
\[ \lambda' = 45.33 \times \frac{17}{18} \approx 42.8 \, \text{cm} \]

### Final Answer
The wavelength of the waves produced in air outside the tube at \( 16°C \) is approximately \( 42.8 \, \text{cm} \).

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