A string is fixed at both end transverse oscillations with amplitude `a_(0)` are excited. Which of the following statements are correct ?

Options

A. (a) Energy of oscillations in the string is directly proportional to tension in the string
B. (b) Energy of oscillations in nth overtone will be equal to `n^(2)` times of that in first overtone
C. (c) Average kinetic energy of string (over an oscillation period) is half of the oscillation energy
D. (d) None of the above

Explanation

To solve the problem regarding the transverse oscillations of a string fixed at both ends, we will analyze each statement provided in the question step by step.

### Step 1: Analyze the Energy of Oscillation in the String
The energy of oscillation in a string can be expressed in terms of its amplitude, tension, and mass per unit length. The formula for the energy (E) in a vibrating string is given by:
\[ E = \frac{1}{2} T A^2 \]
where:
- \( T \) is the tension in the string,
- \( A \) is the amplitude of oscillation.

From this equation, we can see that the energy of oscillation is directly proportional to the tension in the string. Therefore, the first statement is **correct**.

### Step 2: Analyze the Energy in Terms of Overtone
The fundamental frequency (first harmonic) of a string fixed at both ends is given by:
\[ f_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) is the length of the string,
- \( \mu \) is the mass per unit length of the string.

The frequency of the nth overtone (which is the (n+1)th harmonic) can be expressed as:
\[ f_n = (n + 1) f_0 \]
The energy associated with the nth overtone can be expressed as:
\[ E_n \propto f_n^2 \]
This leads us to:
\[ E_n \propto (n + 1)^2 \]
This means that the energy of the nth overtone is proportional to \( (n + 1)^2 \), not \( n^2 \). Therefore, the second statement is **incorrect**.

### Step 3: Analyze the Average Kinetic Energy
The average kinetic energy of the oscillating string can be derived from the total energy of oscillation. The average kinetic energy (K.E.) in a simple harmonic motion is given by:
\[ K.E. = \frac{1}{2} E \]
This implies that the average kinetic energy over one oscillation period is half of the total oscillation energy. Therefore, the third statement is **correct**.

### Conclusion
Based on the analysis:
- The first statement is correct: The energy of oscillation in the string is directly proportional to the tension in the string.
- The second statement is incorrect: The energy in the nth overtone is proportional to \( (n + 1)^2 \), not \( n^2 \).
- The third statement is correct: The average kinetic energy is half of the total oscillation energy.

### Final Answer
The correct statements are:
- A: Correct
- B: Incorrect
- C: Correct