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, we need to find the maximum amplitude of the plank in which the block does not slip over it, given the coefficient of friction, angular frequency, and gravitational acceleration.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Coefficient of friction (μ) = 1/2
- Angular frequency (ω) = 10 rad/s
- Gravitational acceleration (g) = 10 m/s²

2. **Understand the Condition for No Slipping:**
- For the block to not slip over the plank, the maximum static friction must be equal to the pseudo force acting on the block due to the acceleration of the plank.

3. **Calculate Maximum Acceleration of the Plank:**
- The maximum acceleration (a_max) of the plank during SHM can be calculated using the formula:
\[ a_{max} = \omega^2 A \]
- Here, A is the maximum amplitude we need to find.

4. **Set Up the Equation for No Slipping:**
- The force of friction (F_friction) that prevents slipping is given by:
\[ F_{friction} = \mu N \]
- The normal force (N) acting on the block is equal to its weight (mg):
\[ N = mg \]
- Therefore, the friction force can be expressed as:
\[ F_{friction} = \mu mg \]

5. **Relate Friction Force to Pseudo Force:**
- The pseudo force acting on the block due to the acceleration of the plank is:
\[ F_{pseudo} = ma_{max} \]
- For no slipping, we set the friction force equal to the pseudo force:
\[ \mu mg = ma_{max} \]

6. **Cancel Mass (m) from Both Sides:**
- We can cancel mass (m) from both sides of the equation:
\[ \mu g = a_{max} \]

7. **Substitute Values:**
- Substitute the known values into the equation:
\[ \frac{1}{2} \cdot 10 = a_{max} \]
- This gives:
\[ a_{max} = 5 \text{ m/s}^2 \]

8. **Relate Maximum Acceleration to Amplitude:**
- Now, we can relate this back to the maximum amplitude:
\[ a_{max} = \omega^2 A \]
- Rearranging gives:
\[ A = \frac{a_{max}}{\omega^2} \]

9. **Calculate Maximum Amplitude:**
- Substitute the values of \( a_{max} \) and \( \omega \):
\[ A = \frac{5}{10^2} = \frac{5}{100} = 0.05 \text{ m} \]
- Converting to centimeters:
\[ A = 5 \text{ cm} \]

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