The equation of motion of a particle in `SHM` is `a +4x = 0`. Here 'a' is linear acceleration of the particle at displacement 'x' in metre. Its time period is

Options

A. `pis`
B. `2pis`
C. `(pi)/(2)s`
D. `2s`

Explanation

To find the time period of the particle in simple harmonic motion (SHM) given the equation of motion \( a + 4x = 0 \), we can follow these steps:

### Step 1: Rewrite the equation of motion
The given equation is:
\[ a + 4x = 0 \]
We can rearrange this to express acceleration \( a \) in terms of displacement \( x \):
\[ a = -4x \]

### Step 2: Relate the acceleration to angular frequency
In SHM, the acceleration \( a \) can also be expressed in terms of angular frequency \( \omega \) and displacement \( x \) as:
\[ a = -\omega^2 x \]
Now we can compare the two equations we have for acceleration:
\[ -4x = -\omega^2 x \]

### Step 3: Equate the coefficients
Since the equations are equal, we can equate the coefficients of \( x \):
\[ \omega^2 = 4 \]

### Step 4: Solve for \( \omega \)
Taking the square root of both sides gives us:
\[ \omega = 2 \, \text{rad/s} \]

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

### Conclusion
Thus, the time period of the particle in SHM is:
\[ \boxed{\pi \, \text{seconds}} \]

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