A particle executes S.H.M. with a period of 6 second and amplitude of 3 cm . Its maximum speed in cm / sec is

Options

A. `pi//2`
B. `pi`
C. `2pi`
D. `3 pi`

Answer

To find the maximum speed of a particle executing Simple Harmonic Motion (S.H.M.), we can use the following formula:

\[ v_{max} = A \cdot \omega \]

Where:
- \( v_{max} \) is the maximum speed,
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency.

### Step 1: Identify the given values
From the problem, we have:
- Period \( T = 6 \) seconds
- Amplitude \( A = 3 \) cm

### Step 2: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is related to the period \( T \) by the formula:

\[ \omega = \frac{2\pi}{T} \]

Substituting the value of \( T \):

\[ \omega = \frac{2\pi}{6} = \frac{\pi}{3} \, \text{rad/sec} \]

### Step 3: Calculate the maximum speed \( v_{max} \)
Now, we can substitute the values of \( A \) and \( \omega \) into the maximum speed formula:

\[ v_{max} = A \cdot \omega = 3 \, \text{cm} \cdot \frac{\pi}{3} \, \text{rad/sec} \]

This simplifies to:

\[ v_{max} = \pi \, \text{cm/sec} \]

### Final Answer
Thus, the maximum speed of the particle is:

\[ v_{max} \approx 3.14 \, \text{cm/sec} \]