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

Options

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

Answer

To solve the problem, we need to analyze how the maximum acceleration and maximum speed of a particle in Simple Harmonic Motion (SHM) are related to amplitude and frequency.

### Step-by-Step Solution:

1. **Understanding Maximum Speed and Maximum Acceleration in SHM:**
- The maximum speed \( v_{max} \) in SHM is given by the formula:
\[ v_{max} = \omega A \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude.
- The angular frequency \( \omega \) can also be expressed in terms of frequency \( f \) as:
\[ \omega = 2\pi f \]
- Therefore, substituting this into the maximum speed equation gives:
\[ v_{max} = 2\pi f A \]

2. **Understanding Maximum Acceleration in SHM:**
- The maximum acceleration \( a_{max} \) in SHM is given by:
\[ a_{max} = \omega^2 A \]
- Substituting \( \omega = 2\pi f \) into this equation gives:
\[ a_{max} = (2\pi f)^2 A = 4\pi^2 f^2 A \]

3. **Setting Up the Problem:**
- We are told that the maximum acceleration is doubled while keeping the maximum speed constant. Thus:
\[ a'_{max} = 2 a_{max} \]
- This means:
\[ 2(4\pi^2 f^2 A) = 4\pi^2 f^2 A' \]
- We need to find the relationship between the new amplitude \( A' \) and the frequency \( f' \).

4. **Analyzing the Conditions:**
- Since \( v_{max} \) remains constant:
\[ 2\pi f A = 2\pi f' A' \]
- This implies:
\[ f A = f' A' \]

5. **Exploring the Options:**
- **Option 1:** Amplitude is doubled while frequency remains constant.
- \( A' = 2A \), \( f' = f \)
- This leads to \( v_{max} = 2\pi f (2A) = 4\pi f A \) (not constant).
- **Option 2:** Amplitude is doubled while frequency is halved.
- \( A' = 2A \), \( f' = \frac{f}{2} \)
- This leads to \( v_{max} = 2\pi \left(\frac{f}{2}\right)(2A) = 2\pi f A \) (not constant).
- **Option 3:** Frequency is doubled while amplitude is halved.
- \( A' = \frac{A}{2} \), \( f' = 2f \)
- This leads to \( v_{max} = 2\pi (2f) \left(\frac{A}{2}\right) = 2\pi f A \) (constant).
- For acceleration: \( a'_{max} = 4\pi^2 (2f)^2 \left(\frac{A}{2}\right) = 4\pi^2 f^2 A \) (which is double).
- **Option 4:** Frequency is doubled while amplitude remains constant.
- \( A' = A \), \( f' = 2f \)
- This leads to \( v_{max} = 2\pi (2f) A = 4\pi f A \) (not constant).

6. **Conclusion:**
- The only option that satisfies both conditions (maximum speed remains constant and maximum acceleration is doubled) is **Option 3**.

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