A particle executes SHM on a straight line path. The amplitude of oscillation is `2cm`. When the displacement of the particle from the mean position is `1cm`, the numerical value of magnitude of acceleration is equal to the mumerical value of velocity. Find the frequency of SHM (in `Hz`).

Explanation

To solve the problem step by step, we will use the properties of Simple Harmonic Motion (SHM).

### Step 1: Understand the given values
- Amplitude (A) = 2 cm
- Displacement from mean position (x) = 1 cm

### Step 2: Write the equations for acceleration and velocity in SHM
1. The formula for acceleration (a) in SHM is given by:
\[ a = -\omega^2 x \]
where \( \omega \) is the angular frequency and \( x \) is the displacement from the mean position.

2. The formula for velocity (v) in SHM is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]

### Step 3: Set the magnitudes of acceleration and velocity equal
According to the problem, the magnitudes of acceleration and velocity are equal when \( x = 1 \) cm:
\[ |\omega^2 x| = |\omega \sqrt{A^2 - x^2}| \]

### Step 4: Substitute the known values into the equation
Substituting \( A = 2 \) cm and \( x = 1 \) cm into the equation:
\[ \omega^2 (1) = \omega \sqrt{(2^2) - (1^2)} \]
This simplifies to:
\[ \omega^2 = \omega \sqrt{4 - 1} \]
\[ \omega^2 = \omega \sqrt{3} \]

### Step 5: Solve for \( \omega \)
Assuming \( \omega \neq 0 \), we can divide both sides by \( \omega \):
\[ \omega = \sqrt{3} \]

### Step 6: Find the frequency \( f \)
The frequency \( f \) is related to angular frequency \( \omega \) by the formula:
\[ f = \frac{\omega}{2\pi} \]
Substituting \( \omega = \sqrt{3} \):
\[ f = \frac{\sqrt{3}}{2\pi} \]

### Final Answer
The frequency of SHM is:
\[ f = \frac{\sqrt{3}}{2\pi} \text{ Hz} \]

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