`F - x` equation of a body of mass `2kg` in SHM is
`F + 8x = 0`.
Here, `F` is in newton and `x` in meter. Find time period of oscillations.

Options

A. `2.14s`
B. `1s`
C. `3.14s`
D. None of these

Explanation

To find the time period of oscillations for the body in simple harmonic motion (SHM) given the equation \( F + 8x = 0 \), we can follow these steps:

### Step 1: Rewrite the equation
The equation given is:
\[ F + 8x = 0 \]
This can be rearranged to express the force \( F \) in terms of displacement \( x \):
\[ F = -8x \]

### Step 2: Identify the spring constant \( k \)
In SHM, the force can also be expressed as:
\[ F = -kx \]
where \( k \) is the spring constant. By comparing the two equations:
\[ -kx = -8x \]
we can see that:
\[ k = 8 \, \text{N/m} \]

### Step 3: Use the formula for the time period \( T \)
The time period \( T \) of a mass \( m \) attached to a spring (or in SHM) is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass of the body and \( k \) is the spring constant.

### Step 4: Substitute the values
Given that the mass \( m = 2 \, \text{kg} \) and \( k = 8 \, \text{N/m} \), we can substitute these values into the formula:
\[ T = 2\pi \sqrt{\frac{2}{8}} \]

### Step 5: Simplify the expression
Calculating the fraction:
\[ \frac{2}{8} = \frac{1}{4} \]
Now substituting this back into the equation:
\[ T = 2\pi \sqrt{\frac{1}{4}} = 2\pi \cdot \frac{1}{2} = \pi \]

### Step 6: Final result
Thus, the time period \( T \) of the oscillations is:
\[ T \approx 3.14 \, \text{seconds} \]

### Summary of the solution:
The time period of oscillations for the body of mass \( 2 \, \text{kg} \) in SHM is approximately \( 3.14 \, \text{seconds} \).

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