The maximum velocity of a particle performing a S.H.M is v. If the periodic time is made `1//3`rd and the amplitude is doubled, then the new maximum velocity of the particle will be

Options

A. 3 v
B. 6 v
C. `v//6`
D. `3v//2`

Explanation

To solve the problem, we need to determine the new maximum velocity of a particle performing simple harmonic motion (SHM) when the periodic time is made one-third and the amplitude is doubled.

### Step-by-Step Solution:

1. **Understanding Maximum Velocity in SHM**:
The maximum velocity \( V_{\text{max}} \) of a particle in SHM is given by the formula:
\[ V_{\text{max}} = \omega A \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude.

2. **Relating Angular Frequency to Period**:
The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]

3. **Initial Conditions**:
Initially, we have:
- Maximum velocity \( V = \omega A \)
- Time period \( T \)
- Amplitude \( A \)

Therefore, we can express \( V \) as:
\[ V = \left(\frac{2\pi}{T}\right) A \]

4. **New Conditions**:
According to the problem:
- The new time period \( T' = \frac{T}{3} \)
- The new amplitude \( A' = 2A \)

5. **Calculating New Angular Frequency**:
For the new time period:
\[ \omega' = \frac{2\pi}{T'} = \frac{2\pi}{\frac{T}{3}} = \frac{6\pi}{T} \]

6. **Calculating New Maximum Velocity**:
Now, substituting the new values into the maximum velocity formula:
\[ V'_{\text{max}} = \omega' A' = \left(\frac{6\pi}{T}\right)(2A) = \frac{12\pi A}{T} \]

7. **Relating New Maximum Velocity to Original Velocity**:
Recall that:
\[ V = \frac{2\pi A}{T} \]
Therefore, we can express the new maximum velocity in terms of the original maximum velocity:
\[ V'_{\text{max}} = 6 \left(\frac{2\pi A}{T}\right) = 6V \]

### Conclusion:
The new maximum velocity of the particle is:
\[ \boxed{6V} \]