The force of a particle of mass 1 kg is depends on displacement as F = —4x then the frequency of S.H.M. is

Options

A. `pi Hz`
B. `2pi Hz`
C. `1/pi Hz`
D. `1/(2pi) Hz`

Answer

To solve the problem, we need to find the frequency of simple harmonic motion (SHM) for a particle of mass 1 kg, given the force as \( F = -4x \).

### Step-by-Step Solution:

1. **Identify the Force Equation:**
The force acting on the particle is given by:
\[ F = -4x \]

2. **Apply Newton's Second Law:**
According to Newton's second law, force can also be expressed as:
\[ F = m \cdot a \]
where \( m \) is the mass and \( a \) is the acceleration. For our case, the mass \( m = 1 \, \text{kg} \).

3. **Relate Force to Acceleration:**
Setting the two expressions for force equal gives:
\[ m \cdot a = -4x \]
Since \( m = 1 \, \text{kg} \), we have:
\[ a = -4x \]

4. **Express Acceleration in Terms of Displacement:**
The acceleration \( a \) can also be expressed as the second derivative of displacement with respect to time:
\[ a = \frac{d^2x}{dt^2} \]
Therefore, we can rewrite the equation as:
\[ \frac{d^2x}{dt^2} = -4x \]

5. **Identify the Form of the SHM Equation:**
The standard form of the differential equation for SHM is:
\[ \frac{d^2x}{dt^2} + \omega^2 x = 0 \]
Comparing this with our equation, we find:
\[ \omega^2 = 4 \]

6. **Calculate Angular Frequency \( \omega \):**
Taking the square root gives:
\[ \omega = \sqrt{4} = 2 \, \text{rad/s} \]

7. **Relate Angular Frequency to Frequency:**
The relationship between angular frequency \( \omega \) and frequency \( f \) is given by:
\[ \omega = 2\pi f \]
Substituting for \( \omega \):
\[ 2 = 2\pi f \]

8. **Solve for Frequency \( f \):**
Rearranging gives:
\[ f = \frac{2}{2\pi} = \frac{1}{\pi} \, \text{Hz} \]

### Final Answer:
The frequency of the simple harmonic motion is:
\[ f = \frac{1}{\pi} \, \text{Hz} \]