While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of `18 cm` during winter. Repeating the same experiment during summer, she measures the column length to be `x cm` for the second resonance. Then

Options

A. `18 gt x`
B. `x gt 54`
C. `54 gt x gt 36`
D. `36 gt x gt 18`

Answer

To solve the problem, we need to analyze the resonance conditions in a resonance column experiment. Let's break it down step by step:

### Step 1: Understanding the First Resonance Condition
In the winter, the first resonance condition is observed at a column length \( L_1 = 18 \, \text{cm} \). For a column that is open at one end, the first resonance occurs at:

\[ L_1 = \frac{1}{4} \lambda_1 \]

where \( \lambda_1 \) is the wavelength of the sound wave for the first harmonic.

### Step 2: Expressing the Speed of Sound
The speed of sound \( V_1 \) in the winter can be expressed using the relationship between speed, frequency, and wavelength:

\[ V_1 = f_1 \lambda_1 \]

From the first resonance condition, we can express \( \lambda_1 \):

\[ \lambda_1 = 4L_1 = 4 \times 18 \, \text{cm} = 72 \, \text{cm} \]

Thus, we have:

\[ V_1 = f_1 \times 72 \, \text{cm} \]

### Step 3: Analyzing the Second Resonance Condition
In summer, the second resonance condition occurs at a length \( L_2 = x \, \text{cm} \). For the second resonance (which is the third harmonic for a column open at one end), we have:

\[ L_2 = \frac{3}{4} \lambda_2 \]

where \( \lambda_2 \) is the wavelength for the second resonance.

### Step 4: Expressing the Speed of Sound in Summer
The speed of sound \( V_2 \) in summer can be expressed as:

\[ V_2 = f_2 \lambda_2 \]

From the second resonance condition, we can express \( \lambda_2 \):

\[ \lambda_2 = \frac{4}{3} L_2 = \frac{4}{3} x \]

Thus, we have:

\[ V_2 = f_2 \times \frac{4}{3} x \]

### Step 5: Equating the Frequencies
Since the frequencies are equal for both conditions, we can set up the equation:

\[ f_1 = f_2 \]

This leads us to the relationship:

\[ \frac{V_1}{\lambda_1} = \frac{V_2}{\lambda_2} \]

Substituting the expressions for \( V_1 \) and \( V_2 \):

\[ \frac{f_1 \times 72}{72} = \frac{f_2 \times \frac{4}{3} x}{\frac{4}{3} x} \]

### Step 6: Setting Up the Final Equation
From the resonance conditions, we have:

\[ \frac{V_1}{4L_1} = \frac{3V_2}{4L_2} \]

Substituting \( L_1 \) and \( L_2 \):

\[ \frac{V_1}{4 \times 18} = \frac{3V_2}{4x} \]

Cross-multiplying gives:

\[ V_1 \cdot x = 3 \cdot V_2 \cdot 18 \]

Rearranging gives:

\[ x = \frac{54 \cdot V_2}{V_1} \]

### Step 7: Analyzing the Temperature Effect
Since the speed of sound is proportional to the square root of the absolute temperature, and the temperature in summer is higher than in winter, we have:

\[ V_2 > V_1 \]

This implies:

\[ \frac{V_2}{V_1} > 1 \]

### Step 8: Conclusion
Thus, we can conclude:

\[ x = 54 \cdot \frac{V_2}{V_1} > 54 \]

Therefore, the correct option is that \( x > 54 \).

### Final Answer
**Option 2: \( x \) is greater than 54 cm.**

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