A particle of mass `0.1kg` executes `SHM` under a force `F =- 10x (N)`. Speed of particle at mean position is `6 m//s`. Find its amplitude of oscillation.

Options

A. 0.6m
B. 0.2m
C. 0.4m
D. 0.1m

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the given information
We have:
- Mass of the particle, \( m = 0.1 \, \text{kg} \)
- Force acting on the particle, \( F = -10x \, \text{N} \)
- Speed of the particle at the mean position, \( v = 6 \, \text{m/s} \)

### Step 2: Relate force to acceleration
From Newton's second law, we know that:
\[ F = m \cdot a \]
Substituting the expression for force:
\[ -10x = m \cdot a \]
Thus, the acceleration \( a \) can be expressed as:
\[ a = \frac{F}{m} = \frac{-10x}{0.1} = -100x \]

### Step 3: Relate acceleration to angular frequency
In simple harmonic motion (SHM), the acceleration can also be expressed as:
\[ a = -\omega^2 x \]
By comparing the two expressions for acceleration:
\[ -100x = -\omega^2 x \]
We can conclude that:
\[ \omega^2 = 100 \]
Taking the square root gives us:
\[ \omega = 10 \, \text{rad/s} \]

### Step 4: Use the relationship between velocity, amplitude, and angular frequency
At the mean position, the velocity \( v \) is given by the formula:
\[ v = \omega a \]
Substituting the known values:
\[ 6 = 10 \cdot a \]
Solving for \( a \) (amplitude):
\[ a = \frac{6}{10} = 0.6 \, \text{m} \]

### Final Answer
The amplitude of oscillation is:
\[ \boxed{0.6 \, \text{m}} \]

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