Two strings of the same material and the same area of cross-section are used in sonometer experiment. One is loaded with 12kg and the other with 3 kg. The fundamental frequency of the first string is equal to the first overtone of the second string. If the length of the second string is 100 cm, then the length of the first string is

Options

A. 30 cm
B. 200 cm
C. 100 cm
D. 50 cm

Answer

To solve the problem step by step, we will use the relationship between frequency, tension, and length of the strings in a sonometer experiment.

### Step 1: Understand the problem
We have two strings made of the same material and with the same cross-sectional area. One string is loaded with 12 kg and the other with 3 kg. The fundamental frequency of the first string is equal to the first overtone of the second string. We need to find the length of the first string, given that the length of the second string is 100 cm.

### Step 2: Write the formula for frequency
For a string fixed at both ends, the frequency \( f \) is given by the formula:
\[ f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( n \) is the harmonic number,
- \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( \mu \) is the linear mass density of the string.

### Step 3: Define the variables for both strings
Let:
- For the first string (loaded with 12 kg):
- Mass \( m_1 = 12 \, \text{kg} \)
- Length \( L_1 = ? \)
- Frequency \( f_1 \) (fundamental frequency, \( n = 1 \))

- For the second string (loaded with 3 kg):
- Mass \( m_2 = 3 \, \text{kg} \)
- Length \( L_2 = 100 \, \text{cm} = 1 \, \text{m} \)
- Frequency \( f_2 \) (first overtone, \( n = 2 \))

### Step 4: Express the frequencies
For the first string:
\[ f_1 = \frac{1}{2L_1} \sqrt{\frac{m_1 g}{\mu}} = \frac{1}{2L_1} \sqrt{\frac{12 \times 10}{\mu}} \]

For the second string:
\[ f_2 = \frac{2}{2L_2} \sqrt{\frac{m_2 g}{\mu}} = \frac{1}{L_2} \sqrt{\frac{3 \times 10}{\mu}} \]

### Step 5: Set the frequencies equal
Since \( f_1 = f_2 \):
\[ \frac{1}{2L_1} \sqrt{\frac{12 \times 10}{\mu}} = \frac{1}{L_2} \sqrt{\frac{3 \times 10}{\mu}} \]

### Step 6: Simplify the equation
Substituting \( L_2 = 1 \, \text{m} \):
\[ \frac{1}{2L_1} \sqrt{120} = \frac{1}{1} \sqrt{30} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ \sqrt{120} = 2L_1 \sqrt{30} \]

### Step 8: Solve for \( L_1 \)
Squaring both sides:
\[ 120 = 4L_1^2 \cdot 30 \]
\[ 120 = 120L_1^2 \]
\[ L_1^2 = 1 \implies L_1 = 1 \, \text{m} = 100 \, \text{cm} \]

### Final Answer
The length of the first string \( L_1 \) is **100 cm**.

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