The maximum velocity of a particle executing simple harmonic motion is v. If the amplitude is doubled and the time period of oscillation decreased to 1/3 of its original value the maximum velocity becomes

Options

A. 18v
B. 12v
C. 6v
D. 3v

Answer

To solve the problem, we need to find the new maximum velocity of a particle executing simple harmonic motion (SHM) when the amplitude is doubled and the time period is decreased to one-third of its original value.

### Step-by-Step Solution:

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

2. **Relating Angular Frequency to Time Period**:
The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Therefore, we can rewrite the maximum velocity as:
\[ v = A \left(\frac{2\pi}{T}\right) = \frac{2\pi A}{T} \]

3. **Identifying Changes in Amplitude and Time Period**:
According to the problem:
- The amplitude is doubled: \( A' = 2A \)
- The time period is decreased to one-third: \( T' = \frac{T}{3} \)

4. **Calculating New Angular Frequency**:
The new angular frequency \( \omega' \) can be calculated using the new time period:
\[ \omega' = \frac{2\pi}{T'} = \frac{2\pi}{\frac{T}{3}} = \frac{6\pi}{T} \]

5. **Calculating New Maximum Velocity**:
Now, we can find the new maximum velocity \( v' \):
\[ v' = A' \omega' = (2A) \left(\frac{6\pi}{T}\right) = \frac{12\pi A}{T} \]

6. **Relating New Maximum Velocity to Original Maximum Velocity**:
We know that the original maximum velocity \( v = \frac{2\pi A}{T} \). Thus, we can express \( v' \) in terms of \( v \):
\[ v' = 6 \left(\frac{2\pi A}{T}\right) = 6v \]

### Conclusion:
The new maximum velocity \( v' \) becomes \( 6v \).