A resonance air column shows resonance with a tuning fork of frequency 256 Hz at column lengths 33.4 cmand 101.8 cm. find (i) end-correction and (ii) the speed of sound in air.

Options

A. 1. `300 ms^(-1)`
B. 2. `250ms^(-1)`
C. 3. `390ms^(-1)`
D. 4. `350ms^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Resonance Condition
In a resonance air column, the lengths at which resonance occurs correspond to specific harmonics of the sound wave. For a tube closed at one end, the resonant lengths correspond to odd multiples of a quarter wavelength.

### Step 2: Calculate the Wavelength
The difference in lengths at which resonance occurs corresponds to half the wavelength (\( \lambda/2 \)). Given the two lengths:
- \( L_1 = 33.4 \, \text{cm} \)
- \( L_2 = 101.8 \, \text{cm} \)

The difference is:
\[ \Delta L = L_2 - L_1 = 101.8 \, \text{cm} - 33.4 \, \text{cm} = 68.4 \, \text{cm} \]
Since this difference corresponds to \( \lambda/2 \):
\[ \frac{\lambda}{2} = 68.4 \, \text{cm} \implies \lambda = 2 \times 68.4 \, \text{cm} = 136.8 \, \text{cm} \]

### Step 3: Calculate the End Correction
The length of the air column at resonance can be expressed as:
\[ L_n = \frac{n\lambda}{4} + e \]
where \( e \) is the end correction. For the first resonance (n=1):
\[ L_1 = \frac{1 \times \lambda}{4} + e \]
Substituting \( L_1 \) and \( \lambda \):
\[ 33.4 \, \text{cm} = \frac{136.8 \, \text{cm}}{4} + e \]
Calculating \( \frac{136.8}{4} \):
\[ \frac{136.8}{4} = 34.2 \, \text{cm} \]
Now substituting back:
\[ 33.4 \, \text{cm} = 34.2 \, \text{cm} + e \implies e = 33.4 \, \text{cm} - 34.2 \, \text{cm} = -0.8 \, \text{cm} \]

### Step 4: Calculate the Speed of Sound
The speed of sound \( v \) can be calculated using the formula:
\[ v = f \cdot \lambda \]
Given \( f = 256 \, \text{Hz} \) and converting \( \lambda \) to meters:
\[ \lambda = 136.8 \, \text{cm} = 1.368 \, \text{m} \]
Now substituting the values:
\[ v = 256 \, \text{Hz} \times 1.368 \, \text{m} = 349.248 \, \text{m/s} \]
Rounding this, we find:
\[ v \approx 350 \, \text{m/s} \]

### Final Answers
(i) End Correction, \( e = -0.8 \, \text{cm} \)
(ii) Speed of Sound, \( v \approx 350 \, \text{m/s} \)

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