A block is resting on a piston which executes simple harmonic motion in vertical plain with a period of `2.0s` in vertical plane at an amplitude just sufficient for the block to separate from the piston. The maximum velocity of the piston is

Options

A. `(5)/(pi)(m)/(s)`
B. `(10)/(pi)(m)/(s)`
C. `(pi)/(2)(m)/(s)`
D. `(20)/(pi)(m)/(s)`

Explanation

To solve the problem, we need to find the maximum velocity of a piston executing simple harmonic motion (SHM) with a given period. Let’s break down the solution step by step.

### Step 1: Understanding the Problem
The piston executes SHM in a vertical plane with a period \( T = 2.0 \, \text{s} \). The amplitude is just sufficient for the block to separate from the piston, which means that at the maximum displacement, the block loses contact with the piston.

### Step 2: Finding Angular Frequency (\( \omega \))
The angular frequency \( \omega \) is related to the period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the given period:
\[ \omega = \frac{2\pi}{2.0} = \pi \, \text{rad/s} \]

### Step 3: Maximum Acceleration (\( a_{\text{max}} \))
In SHM, the maximum acceleration is given by:
\[ a_{\text{max}} = A \omega^2 \]
where \( A \) is the amplitude. However, since the block is about to separate from the piston, the maximum acceleration will equal the acceleration due to gravity \( g \):
\[ a_{\text{max}} = g \]
Thus, we have:
\[ g = A \omega^2 \]

### Step 4: Finding Amplitude (\( A \))
Rearranging the equation for amplitude \( A \):
\[ A = \frac{g}{\omega^2} \]

### Step 5: Maximum Velocity (\( V_{\text{max}} \))
The maximum velocity in SHM is given by:
\[ V_{\text{max}} = A \omega \]
Substituting the expression for \( A \):
\[ V_{\text{max}} = \left(\frac{g}{\omega^2}\right) \omega = \frac{g}{\omega} \]

### Step 6: Substitute Values
Now substituting \( g = 10 \, \text{m/s}^2 \) and \( \omega = \pi \, \text{rad/s} \):
\[ V_{\text{max}} = \frac{10}{\pi} \]

### Final Answer
Thus, the maximum velocity of the piston is:
\[ V_{\text{max}} \approx \frac{10}{3.14} \approx 3.18 \, \text{m/s} \]