The the resonance tube experiment first resonant length is `l_(1)` and the second resonant length is `l_(2)`, then the third resonant length will be ?

Options

A. `5l`
B. `2(l_(2)-l_(1))`
C. `2l_(2)-l_(1)`
D. `3l_(2)-2l_(1)`

Explanation

To find the third resonant length \( L_3 \) in the resonance tube experiment given the first resonant length \( L_1 \) and the second resonant length \( L_2 \), we can follow these steps:

### Step 1: Understand the Resonance Condition
In a resonance tube, the resonant lengths correspond to the positions where standing waves are formed. The first resonant length \( L_1 \) corresponds to the first harmonic (fundamental frequency), the second resonant length \( L_2 \) corresponds to the third harmonic, and the third resonant length \( L_3 \) will correspond to the fifth harmonic.

### Step 2: Write the Equations for Resonant Lengths
The resonant lengths can be expressed as:
1. For the first resonance:
\[ L_1 + \epsilon = \frac{V}{4f_0} \]
2. For the second resonance:
\[ L_2 + \epsilon = \frac{3V}{4f_0} \]
3. For the third resonance:
\[ L_3 + \epsilon = \frac{5V}{4f_0} \]

Where \( \epsilon \) is the end correction, \( V \) is the speed of sound in air, and \( f_0 \) is the fundamental frequency.

### Step 3: Rearranging the Equations
From the above equations, we can express \( L_1 \), \( L_2 \), and \( L_3 \) in terms of \( \epsilon \):
1. \( L_1 = \frac{V}{4f_0} - \epsilon \)
2. \( L_2 = \frac{3V}{4f_0} - \epsilon \)
3. \( L_3 = \frac{5V}{4f_0} - \epsilon \)

### Step 4: Eliminate \( \epsilon \)
To find \( L_3 \) in terms of \( L_1 \) and \( L_2 \), we can eliminate \( \epsilon \):
- From the equation for \( L_1 \):
\[ \epsilon = \frac{V}{4f_0} - L_1 \]
- Substitute \( \epsilon \) into the equation for \( L_2 \):
\[ L_2 = \frac{3V}{4f_0} - \left(\frac{V}{4f_0} - L_1\right) \]
Simplifying this gives:
\[ L_2 = \frac{3V}{4f_0} - \frac{V}{4f_0} + L_1 = \frac{2V}{4f_0} + L_1 = \frac{V}{2f_0} + L_1 \]

### Step 5: Find \( L_3 \)
Now, substitute \( \epsilon \) into the equation for \( L_3 \):
\[ L_3 = \frac{5V}{4f_0} - \left(\frac{V}{4f_0} - L_1\right) \]
This simplifies to:
\[ L_3 = \frac{5V}{4f_0} - \frac{V}{4f_0} + L_1 = \frac{4V}{4f_0} + L_1 = L_1 + \frac{V}{f_0} \]

### Step 6: Relate \( L_1 \) and \( L_2 \)
From the earlier relationship:
\[ L_2 = L_1 + \frac{V}{2f_0} \]
We can express \( L_3 \) in terms of \( L_1 \) and \( L_2 \):
\[ L_3 = 2L_2 - L_1 \]

### Final Result
Thus, the third resonant length \( L_3 \) can be expressed as:
\[ L_3 = 2L_2 - L_1 \]