The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 6.0m. If the piston moves with simple harmonic motion with an angular frequency of `200 (rad)/min`, what is its maximum speed ?

Options

A. `5ms^(-1)`
B. `10ms^(-1)`
C. `15ms^-1`
D. `20ms^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the amplitude of the motion
The stroke of the piston is given as 6.0 m, and it is stated that the stroke is twice the amplitude. Therefore, we can calculate the amplitude (A) as follows:

\[ A = \frac{\text{Stroke}}{2} = \frac{6.0 \, \text{m}}{2} = 3.0 \, \text{m} \]

### Step 2: Convert angular frequency to radians per second
The angular frequency (ω) is given as 200 rad/min. To convert this to radians per second, we use the conversion factor of 60 seconds in a minute:

\[ \omega = 200 \, \text{rad/min} \times \frac{1 \, \text{min}}{60 \, \text{s}} = \frac{200}{60} \, \text{rad/s} \approx 3.33 \, \text{rad/s} \]

### Step 3: Calculate the maximum speed
The maximum speed (V_max) in simple harmonic motion is given by the formula:

\[ V_{\text{max}} = A \cdot \omega \]

Substituting the values we have:

\[ V_{\text{max}} = 3.0 \, \text{m} \times \frac{200}{60} \, \text{rad/s} = 3.0 \, \text{m} \times 3.33 \, \text{rad/s} \approx 10.0 \, \text{m/s} \]

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

\[ V_{\text{max}} = 10.0 \, \text{m/s} \]