A block rides on position that is moving vertically with simple harmonic motion. The maximum speed of the piston is 2m/s. At what amplitude of motion will the block and piston separate? (g=10m/`s^(2)`)

Options

A. 20 cm
B. 30cm
C. 40 cm
D. 50 cm

Answer

To solve the problem, we need to determine the amplitude of motion at which the block and the piston will separate. We will use the relationship between the maximum speed of the piston, the angular frequency, and the amplitude of the simple harmonic motion.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Maximum speed of the piston, \( v_{max} = 2 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Relate Maximum Speed to Angular Frequency and Amplitude**:
The maximum speed \( v_{max} \) in simple harmonic motion is given by the formula:
\[ v_{max} = \omega A \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude.

3. **Express Angular Frequency**:
The angular frequency \( \omega \) can be related to the acceleration due to gravity \( g \) and the amplitude \( A \) at the point of separation:
\[ \omega^2 A = g \]

4. **Substituting for Angular Frequency**:
From the first equation, we can express \( \omega \) as:
\[ \omega = \frac{v_{max}}{A} \]
Substituting this into the second equation gives:
\[ \left(\frac{v_{max}}{A}\right)^2 A = g \]
Simplifying this, we have:
\[ \frac{v_{max}^2}{A} = g \]

5. **Solve for Amplitude \( A \)**:
Rearranging the equation to solve for \( A \):
\[ A = \frac{v_{max}^2}{g} \]
Now substituting the known values:
\[ A = \frac{(2 \, \text{m/s})^2}{10 \, \text{m/s}^2} = \frac{4}{10} = 0.4 \, \text{m} \]

6. **Convert to Centimeters**:
To express the amplitude in centimeters:
\[ A = 0.4 \, \text{m} = 40 \, \text{cm} \]

### Final Answer:
The amplitude of motion at which the block and piston will separate is **40 cm**.