A block is kept on a rough horizontal plank. The coefficient of friction between the block and the plank is `1/2`. The plank is undergoing SHM of angular frequency 10 rad/s. The maximum amplitude of plank in which the block does not slip over the plank is (g= 10 m/`s^(2)`)

Options

A. 4cm
B. 5cm
C. 10 cm
D. 16 cm

Answer

To solve the problem step by step, we will analyze the forces acting on the block and the conditions under which it does not slip on the plank undergoing simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Identify the Forces**:
- The block experiences a gravitational force \( mg \) acting downwards.
- The normal force \( N \) acts upwards, which is equal to \( mg \) since there is no vertical motion.
- The frictional force \( f \) that prevents the block from slipping is given by \( f = \mu N = \mu mg \), where \( \mu \) is the coefficient of friction.

2. **Acceleration Due to SHM**:
- The plank is undergoing SHM with angular frequency \( \omega = 10 \, \text{rad/s} \).
- The acceleration \( a \) of the plank at maximum displacement (amplitude \( x \)) is given by:
\[ a = -\omega^2 x \]
- The negative sign indicates that the acceleration is directed towards the mean position.

3. **Condition for No Slipping**:
- For the block to not slip, the maximum pseudo force (which is the inertial force due to the acceleration of the plank) must be less than or equal to the maximum frictional force:
\[ f_{\text{max}} = \mu mg \]
- The pseudo force acting on the block due to the SHM of the plank is:
\[ F_{\text{pseudo}} = m a = m (-\omega^2 x) \]

4. **Setting Up the Equation**:
- For the block to remain at rest relative to the plank, we set the magnitudes of the forces equal:
\[ \mu mg = m \omega^2 x \]
- We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu g = \omega^2 x \]

5. **Solving for Maximum Amplitude \( x \)**:
- Rearranging the equation gives:
\[ x = \frac{\mu g}{\omega^2} \]

6. **Substituting Values**:
- Given \( \mu = \frac{1}{2} \), \( g = 10 \, \text{m/s}^2 \), and \( \omega = 10 \, \text{rad/s} \):
\[ x = \frac{\frac{1}{2} \cdot 10}{10^2} = \frac{5}{100} = \frac{1}{20} \, \text{m} \]

7. **Final Answer**:
- Converting to centimeters:
\[ x = \frac{1}{20} \, \text{m} = 5 \, \text{cm} \]

### Conclusion:
The maximum amplitude of the plank in which the block does not slip over it is **5 cm**.