A cubical block of mass M vibrates horizontally with amplitude of 4.0 cm and a frequency of 2.0 Hz. A small block of mass is placed on the bigger block. In order that the smaller block does not side on the bigger block, the minimum value of the coefficient of static friction between the two blocks is (0.16x). Find the valule of x(Take `pi^(2)`=10 and g=10m/`s^(2)`)

Options

A. 0.36
B. 0.4
C. 0.64
D. 0.72

Explanation

To solve the problem, we need to find the minimum value of the coefficient of static friction (\( \mu \)) between the two blocks so that the smaller block does not slide off the larger block. We will use the relationship between the maximum acceleration of the block in simple harmonic motion (SHM) and the gravitational force acting on the smaller block.

### Step-by-step Solution:

1. **Identify the parameters:**
- Amplitude \( A = 4.0 \, \text{cm} = 0.04 \, \text{m} \)
- Frequency \( f = 2.0 \, \text{Hz} \)
- Gravitational acceleration \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the angular frequency (\( \omega \)):**
\[ \omega = 2 \pi f \]
Substituting the frequency:
\[ \omega = 2 \times 3.14 \times 2 = 12.56 \, \text{rad/s} \]

3. **Calculate the maximum acceleration (\( a_{\text{max}} \)) in SHM:**
The maximum acceleration is given by:
\[ a_{\text{max}} = \omega^2 A \]
Substituting the values:
\[ a_{\text{max}} = (12.56)^2 \times 0.04 \]
\[ a_{\text{max}} = 158.1136 \times 0.04 = 6.324544 \, \text{m/s}^2 \]

4. **Set up the inequality for static friction:**
For the smaller block not to slide, the maximum static friction must be greater than or equal to the inertial force acting on it due to the acceleration:
\[ \mu mg \geq ma_{\text{max}} \]
Dividing both sides by \( m \) (mass of the smaller block):
\[ \mu g \geq a_{\text{max}} \]

5. **Rearranging the inequality:**
\[ \mu \geq \frac{a_{\text{max}}}{g} \]

6. **Substituting the values:**
\[ \mu \geq \frac{6.324544}{10} = 0.6324544 \]

7. **Finding the value of \( x \):**
We are given that \( \mu = 0.16x \). Therefore:
\[ 0.16x \geq 0.6324544 \]
Solving for \( x \):
\[ x \geq \frac{0.6324544}{0.16} \approx 3.96535 \]
Rounding to two decimal places, we find:
\[ x \approx 3.97 \]

### Final Answer:
The value of \( x \) is approximately \( 3.97 \).