`F-x` equation of a body in SHM is
`F + 4x=0`
Here, `F` is in newton and `x` in meter. Mass of the body is `1 kg`. Find time period of oscillations.

Explanation

To find the time period of oscillations for a body in Simple Harmonic Motion (SHM) given the force-displacement equation \( F + 4x = 0 \), we can follow these steps:

### Step 1: Rewrite the Force Equation
The equation given is:
\[ F + 4x = 0 \]
This can be rearranged to express \( F \) as:
\[ F = -4x \]

### Step 2: Relate Force to Mass and Acceleration
According to Newton's second law, the force can also be expressed as:
\[ F = ma \]
where \( m \) is the mass and \( a \) is the acceleration. Given that the mass \( m = 1 \, \text{kg} \), we can substitute this into the equation:
\[ ma = -4x \]
This simplifies to:
\[ 1 \cdot a = -4x \quad \Rightarrow \quad a = -4x \]

### Step 3: Relate Acceleration to SHM
In SHM, the acceleration \( a \) is also given by:
\[ a = -\omega^2 x \]
where \( \omega \) is the angular frequency. Setting the two expressions for acceleration equal to each other gives:
\[ -4x = -\omega^2 x \]

### Step 4: Solve for Angular Frequency
Since \( x \) is not zero, we can divide both sides by \( x \):
\[ 4 = \omega^2 \]
Taking the square root of both sides, we find:
\[ \omega = 2 \, \text{radians/second} \]

### Step 5: Calculate the Time Period
The time period \( T \) of oscillation is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{2} = \pi \, \text{seconds} \]

### Final Answer
Thus, the time period of the oscillations is:
\[ T = \pi \, \text{seconds} \approx 3.14 \, \text{seconds} \]

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