A particle executes linear simple harmonic motion with an amplitude of `3 cm`. When the particle is at `2 cm` from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then, its time period in seconds is

Options

A. `(sqrt(5))/(pi)`
B. `(sqrt(5))/(2pi)`
C. `(4pi)/(sqrt(4))`
D. `(2pi)/(sqrt(3))`

Answer

To solve the problem, we need to find the time period of a particle executing simple harmonic motion (SHM) given certain conditions. Let's break this down step by step.

### Step 1: Understand the parameters of SHM
The amplitude \( A \) of the SHM is given as \( 3 \, \text{cm} \). The particle is at a displacement \( x = 2 \, \text{cm} \) from the mean position.

### Step 2: Write the equations for velocity and acceleration in SHM
In SHM, the velocity \( v \) and acceleration \( a \) can be expressed as:
- Velocity:
\[ v = \omega \sqrt{A^2 - x^2} \]
- Acceleration:
\[ a = -\omega^2 x \]
Here, \( \omega \) is the angular frequency.

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

### Step 4: Substitute the values
Substituting \( A = 3 \, \text{cm} \) and \( x = 2 \, \text{cm} \):
\[ \omega \sqrt{3^2 - 2^2} = \omega^2 \cdot 2 \]
This simplifies to:
\[ \omega \sqrt{9 - 4} = 2\omega^2 \]
\[ \omega \sqrt{5} = 2\omega^2 \]

### Step 5: Solve for \( \omega \)
Assuming \( \omega \neq 0 \), we can divide both sides by \( \omega \):
\[ \sqrt{5} = 2\omega \]
Thus,
\[ \omega = \frac{\sqrt{5}}{2} \]

### Step 6: Calculate the time period \( T \)
The time period \( T \) is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{\frac{\sqrt{5}}{2}} = \frac{4\pi}{\sqrt{5}} \]

### Final Answer
The time period of the particle in seconds is:
\[ T = \frac{4\pi}{\sqrt{5}} \, \text{seconds} \]

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