The equation of S.H.M. of a particle is `a+4pi^(2)x=0`, where a is instantaneous linear acceleration at displacement x. Then the frequency of motion is

Options

A. 1 Hz
B. `4pi Hz`
C. `1/4 Hz`
D. 4 Hz

Answer

To solve the problem, we need to analyze the given equation of simple harmonic motion (S.H.M.) and derive the frequency from it. The equation provided is:

\[ a + 4\pi^2 x = 0 \]

where \( a \) is the instantaneous linear acceleration and \( x \) is the displacement from the mean position.

### Step 1: Identify the relationship between acceleration and displacement
In S.H.M., the acceleration \( a \) is related to the displacement \( x \) by the formula:

\[ a = -\omega^2 x \]

where \( \omega \) is the angular frequency of the motion.

### Step 2: Rearranging the given equation
From the given equation, we can express the acceleration in terms of displacement:

\[ a = -4\pi^2 x \]

### Step 3: Equating the two expressions for acceleration
Now, we can equate the two expressions for acceleration:

\[ -\omega^2 x = -4\pi^2 x \]

### Step 4: Solving for angular frequency \( \omega \)
By comparing the coefficients of \( x \), we find:

\[ \omega^2 = 4\pi^2 \]

Taking the square root of both sides gives:

\[ \omega = 2\pi \]

### Step 5: Finding the frequency \( f \)
The frequency \( f \) is related to the angular frequency \( \omega \) by the formula:

\[ f = \frac{\omega}{2\pi} \]

Substituting the value of \( \omega \):

\[ f = \frac{2\pi}{2\pi} = 1 \, \text{Hz} \]

### Conclusion
Thus, the frequency of the motion is:

\[ \boxed{1 \, \text{Hz}} \]

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