A block on a horizontal slab is moving horizontally with a simple harmonic motion of frequency two oscillations per second. If the coefficient of static friction between block and slab is 0.5, then the amplitude of the oscillation will be (if the block does not slip along the slab)

Options

A. 3.3 cm
B. 3.5 cm
C. 3.15 cm
D. 7 cm

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript:

### Step 1: Understand the Problem
We have a block on a horizontal slab that is undergoing simple harmonic motion (SHM) with a frequency of 2 oscillations per second. The coefficient of static friction between the block and the slab is given as 0.5. We need to find the maximum amplitude of the oscillation such that the block does not slip on the slab.

**Hint:** Identify the key parameters: frequency, coefficient of static friction, and the relationship between force and motion.

### Step 2: Identify Forces Acting on the Block
In SHM, the restoring force acting on the block can be described by Hooke's Law, which states that the force is proportional to the displacement from the equilibrium position. The force exerted by the spring can be expressed as:
\[ F = kx \]
where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

The frictional force that prevents slipping is given by:
\[ F_{\text{friction}} = \mu_s mg \]
where \( \mu_s \) is the coefficient of static friction, \( m \) is the mass of the block, and \( g \) is the acceleration due to gravity.

**Hint:** Write down the expressions for the forces involved.

### Step 3: Set Up the Equation for No Slipping Condition
For the block to not slip, the maximum static friction must be equal to the maximum restoring force at the amplitude \( A \):
\[ \mu_s mg = kA \]

**Hint:** Relate the forces using the condition for static friction.

### Step 4: Relate Spring Constant to Angular Frequency
The spring constant \( k \) can also be related to the angular frequency \( \omega \) of the oscillation:
\[ k = m\omega^2 \]
where \( \omega = 2\pi f \).

**Hint:** Use the relationship between frequency and angular frequency.

### Step 5: Substitute and Rearrange
Substituting \( k \) into the friction equation gives:
\[ \mu_s mg = m\omega^2 A \]
Canceling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu_s g = \omega^2 A \]
Now, we can solve for \( A \):
\[ A = \frac{\mu_s g}{\omega^2} \]

**Hint:** Rearrange the equation to isolate \( A \).

### Step 6: Calculate Angular Frequency
Given the frequency \( f = 2 \) Hz, we can calculate \( \omega \):
\[ \omega = 2\pi f = 2\pi \times 2 = 4\pi \]
Thus,
\[ \omega^2 = (4\pi)^2 = 16\pi^2 \]

**Hint:** Calculate \( \omega \) using the frequency provided.

### Step 7: Substitute Values
Now substitute \( \mu_s = 0.5 \), \( g = 10 \, \text{m/s}^2 \), and \( \omega^2 = 16\pi^2 \) into the equation for \( A \):
\[ A = \frac{0.5 \times 10}{16\pi^2} \]
\[ A = \frac{5}{16\pi^2} \]

### Step 8: Calculate Amplitude
Now, we can compute the numerical value of \( A \):
\[ A \approx \frac{5}{16 \times 9.87} \approx \frac{5}{158} \approx 0.0316 \, \text{m} \]
Converting to centimeters:
\[ A \approx 3.16 \, \text{cm} \]

**Hint:** Perform the final calculation to find the amplitude in centimeters.

### Final Answer
The amplitude of the oscillation will be approximately **3.16 cm**.