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

Options

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

Answer

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

### Step 1: Identify the given parameters
- 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:
\[ F = ma \]
Substituting the expression for force:
\[ -10x = m \cdot a \]
Since acceleration \( a \) can also be expressed as \( a = \frac{F}{m} \):
\[ a = \frac{-10x}{0.1} = -100x \]

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

### Step 4: Use the relationship between velocity, angular frequency, and amplitude
At the mean position, the velocity \( v \) is given by:
\[ v = \omega A \]
Where \( A \) is the amplitude. We can rearrange this to find \( A \):
\[ A = \frac{v}{\omega} \]
Substituting the known values:
\[ A = \frac{6 \, \text{m/s}}{10 \, \text{rad/s}} = 0.6 \, \text{m} \]

### Step 5: Conclusion
The amplitude of the oscillations is:
\[ \boxed{0.6 \, \text{m}} \]

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