The maximum acceleration of a particle un SHM is made two times keeping the maximum speed to be constant. It is possible when .

Options

A. amplitude of oscillation is double while frequency remains constant
B. amplitude is double while frequency is halved
C. frequency is doubled while amplitude is halved
D. frequency of oscillation is doubled while amplitude remians constant

Explanation

To solve the problem, we need to analyze the relationships between maximum acceleration, maximum speed, amplitude, and frequency in Simple Harmonic Motion (SHM).

### Step-by-Step Solution:

1. **Understanding Maximum Acceleration and Speed in SHM**:
- The maximum acceleration \( a_{max} \) in SHM is given by the formula:
\[ a_{max} = \omega^2 A \]
- The maximum speed \( v_{max} \) is given by:
\[ v_{max} = \omega A \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude.

2. **Setting Up the Problem**:
- We want to double the maximum acceleration while keeping the maximum speed constant. This means:
\[ a'_{max} = 2a_{max} \]
- Since \( v_{max} \) must remain constant, we can express this as:
\[ v'_{max} = v_{max} \]

3. **Analyzing the Options**:
- We will evaluate each option to see if it satisfies the conditions of doubling the maximum acceleration while keeping the maximum speed constant.

4. **Option 1: Amplitude is doubled while frequency remains constant**:
- If \( A \) is doubled, then:
\[ a'_{max} = \omega^2 (2A) = 2a_{max} \]
- However, \( v'_{max} = \omega (2A) = 2v_{max} \), which does not keep \( v_{max} \) constant. **This option is incorrect.**

5. **Option 2: Amplitude is doubled while frequency is halved**:
- If \( A \) is doubled and frequency is halved (\( \omega \) becomes \( \frac{\omega}{2} \)):
\[ a'_{max} = \left(\frac{\omega}{2}\right)^2 (2A) = \frac{\omega^2}{4} (2A) = \frac{\omega^2 A}{2} = \frac{a_{max}}{2} \]
- Here, \( v'_{max} = \frac{\omega}{2} (2A) = \omega A = v_{max} \) remains constant. However, \( a'_{max} \) is not doubled. **This option is incorrect.**

6. **Option 3: Frequency is doubled while amplitude is halved**:
- If frequency is doubled (\( \omega \) becomes \( 2\omega \)) and amplitude is halved (\( A \) becomes \( \frac{A}{2} \)):
\[ a'_{max} = (2\omega)^2 \left(\frac{A}{2}\right) = 4\omega^2 \left(\frac{A}{2}\right) = 2\omega^2 A = 2a_{max} \]
- For maximum speed:
\[ v'_{max} = (2\omega) \left(\frac{A}{2}\right) = \omega A = v_{max} \]
- Both conditions are satisfied: maximum acceleration is doubled and maximum speed remains constant. **This option is correct.**

7. **Option 4: Frequency is doubled while amplitude remains constant**:
- If frequency is doubled and amplitude remains constant:
\[ a'_{max} = (2\omega)^2 A = 4\omega^2 A = 4a_{max} \]
- Here, \( v'_{max} = 2\omega A \), which does not keep \( v_{max} \) constant. **This option is incorrect.**

### Conclusion:
The correct option is **Option 3**: Frequency is doubled while amplitude is halved.