A `20 cm` long rubber string fixed at both ends obeys Hook's law. Intially when it is stretched to make its total length of `24 cm`, the lowest frequency resonance is `v_(0)`. It is further stretched to make its total length of `26 cm`. The lowest frequency of resonance will now be :

Options

A. the same as `v_(0)`
B. greter than `v_(0)`
C. lower than `v_(0)`
D. None of these

Explanation

To solve the problem, we need to analyze the relationship between the frequency of resonance of a stretched string and its tension and length. Let's break it down step by step.

### Step 1: Understand the formula for frequency of resonance
The lowest frequency of resonance \( f \) for a string fixed at both ends is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( \mu \) is the mass per unit length of the string.

### Step 2: Analyze the initial condition
Initially, the string is stretched to a length of \( L_1 = 24 \, \text{cm} = 0.24 \, \text{m} \). The lowest frequency of resonance at this length is denoted as \( f_0 \):
\[ f_0 = \frac{1}{2 \times 0.24} \sqrt{\frac{T_1}{\mu}} \]

### Step 3: Analyze the second condition
The string is then stretched to a length of \( L_2 = 26 \, \text{cm} = 0.26 \, \text{m} \). The new frequency of resonance \( f_1 \) can be expressed as:
\[ f_1 = \frac{1}{2 \times 0.26} \sqrt{\frac{T_2}{\mu}} \]

### Step 4: Determine the change in tension
When the string is stretched from \( L_1 \) to \( L_2 \), the tension \( T \) in the string increases because the string is being stretched further. Thus, we can say:
\[ T_2 > T_1 \]

### Step 5: Compare the frequencies
Since the frequency is inversely proportional to the length of the string and directly proportional to the square root of the tension, we can conclude:
- As the length increases from \( L_1 \) to \( L_2 \), the term \( \frac{1}{2L} \) decreases.
- As the tension increases from \( T_1 \) to \( T_2 \), the term \( \sqrt{T} \) increases.

Therefore, the overall effect on frequency \( f_1 \) compared to \( f_0 \) is:
\[ f_1 > f_0 \]

### Conclusion
The lowest frequency of resonance when the string is stretched to \( 26 \, \text{cm} \) will be greater than the initial frequency \( f_0 \).

### Final Answer
The lowest frequency of resonance will now be **greater than** \( v_0 \).

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