A body is on a rough horizontal surface which is moving horizontally in SHM of frequency `2Hz`. The coefficient of static friction between the body and the surface is `0.5`. The maximum value of the amplitude for which the body will not slip along the surface is approximately

Options

A. `9cm`
B. `6cm`
C. `4.5cm`
D. `3cm`

Explanation

To solve the problem, we need to find the maximum amplitude \( A \) for which the body will not slip on the rough horizontal surface that is undergoing simple harmonic motion (SHM) with a frequency of \( 2 \, \text{Hz} \). The coefficient of static friction between the body and the surface is given as \( 0.5 \).

### Step-by-Step Solution:

1. **Identify Given Values**:
- Frequency \( f = 2 \, \text{Hz} \)
- Coefficient of static friction \( \mu = 0.5 \)
- Acceleration due to gravity \( g \approx 9.81 \, \text{m/s}^2 \)

2. **Calculate Angular Frequency**:
The angular frequency \( \omega \) can be calculated using the formula:
\[ \omega = 2 \pi f \]
Substituting the given frequency:
\[ \omega = 2 \pi \times 2 = 4 \pi \, \text{rad/s} \]

3. **Determine Maximum Acceleration**:
The maximum acceleration \( a_{\text{max}} \) of the body on the surface can be expressed in terms of amplitude \( A \):
\[ a_{\text{max}} = \omega^2 A \]
Substituting \( \omega \):
\[ a_{\text{max}} = (4 \pi)^2 A = 16 \pi^2 A \]

4. **Friction Force and Condition for No Slipping**:
The maximum static friction force \( F_{\text{friction}} \) that can act on the body is given by:
\[ F_{\text{friction}} = \mu N = \mu mg \]
where \( N = mg \) is the normal force acting on the body.

5. **Set Up the Equation for No Slipping**:
For the body to not slip, the maximum static friction must be equal to the required force to keep the body accelerating:
\[ \mu mg = ma_{\text{max}} \]
Substituting \( a_{\text{max}} \):
\[ \mu mg = m(16 \pi^2 A) \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu g = 16 \pi^2 A \]

6. **Solve for Amplitude \( A \)**:
Rearranging the equation gives:
\[ A = \frac{\mu g}{16 \pi^2} \]
Substituting the values of \( \mu \) and \( g \):
\[ A = \frac{0.5 \times 9.81}{16 \pi^2} \]

7. **Calculate the Numerical Value**:
First, calculate \( 16 \pi^2 \):
\[ 16 \pi^2 \approx 16 \times 9.87 \approx 157.92 \]
Now substituting into the equation for \( A \):
\[ A \approx \frac{4.905}{157.92} \approx 0.0311 \, \text{m} \]

8. **Convert to Centimeters**:
To convert meters to centimeters:
\[ A \approx 0.0311 \, \text{m} \times 100 \approx 3.11 \, \text{cm} \]

### Final Answer:
The maximum value of the amplitude for which the body will not slip along the surface is approximately \( 3.11 \, \text{cm} \).