The amplitude of a particle executing SHM is made three-fourth keeping its time period constant. Its total energy will be

Options

A. `E/2`
B. `3/4 E`
C. `9/16 E`
D. none of these

Answer

To solve the problem, we need to understand how the total energy of a particle executing Simple Harmonic Motion (SHM) is related to its amplitude and time period.

### Step-by-step Solution:

1. **Understand the formula for total energy in SHM**:
The total energy (E) of a particle in SHM is given by the formula:
\[ E = \frac{1}{2} k A^2 \]
where \( k \) is the spring constant and \( A \) is the amplitude of the motion.

2. **Identify the initial conditions**:
Let the initial amplitude be \( A \) and the total energy be \( E \). Therefore, we can express the initial energy as:
\[ E = \frac{1}{2} k A^2 \]

3. **Change the amplitude**:
The problem states that the amplitude is changed to three-fourths of its original value. Thus, the new amplitude \( A' \) is:
\[ A' = \frac{3}{4} A \]

4. **Calculate the new total energy**:
Now, substituting \( A' \) into the energy formula, the new total energy \( E' \) becomes:
\[ E' = \frac{1}{2} k (A')^2 = \frac{1}{2} k \left(\frac{3}{4} A\right)^2 \]
Simplifying this gives:
\[ E' = \frac{1}{2} k \left(\frac{9}{16} A^2\right) = \frac{9}{16} \left(\frac{1}{2} k A^2\right) = \frac{9}{16} E \]

5. **Conclusion**:
Therefore, the new total energy \( E' \) is:
\[ E' = \frac{9}{16} E \]

### Final Answer:
The total energy will be \( \frac{9E}{16} \).

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