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(5))`
D. `(2pi)/(sqrt(3))`

Answer

To solve the problem step by step, we will use the formulas related to simple harmonic motion (SHM).

### Step 1: Understand the problem
We have a particle executing linear simple harmonic motion with an amplitude \( A = 3 \, \text{cm} \). When the particle is at a distance \( x = 2 \, \text{cm} \) from the mean position, the magnitude of its velocity \( v \) is equal to the magnitude of its acceleration \( a \).

### Step 2: Write the formulas for velocity and acceleration
In SHM, the formulas for velocity \( v \) and acceleration \( a \) are:
- 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, we set the magnitudes equal:
\[ |\omega \sqrt{A^2 - x^2}| = |\omega^2 x| \]

### Step 4: Simplify the equation
Since \( \omega \) is positive, we can cancel it from both sides:
\[ \sqrt{A^2 - x^2} = \omega x \]

### Step 5: Substitute the known values
We know \( A = 3 \, \text{cm} \) and \( x = 2 \, \text{cm} \):
\[ \sqrt{3^2 - 2^2} = \omega \cdot 2 \]
Calculating the left side:
\[ \sqrt{9 - 4} = \sqrt{5} \]
So we have:
\[ \sqrt{5} = 2\omega \]

### Step 6: Solve for \( \omega \)
Rearranging gives:
\[ \omega = \frac{\sqrt{5}}{2} \]

### Step 7: Find the time period \( T \)
The time period \( T \) is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting \( \omega \):
\[ T = \frac{2\pi}{\frac{\sqrt{5}}{2}} = \frac{4\pi}{\sqrt{5}} \]

### Final Answer
The time period \( T \) is:
\[ T = \frac{4\pi}{\sqrt{5}} \, \text{seconds} \]

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