A particle executes linear simple harmonic motion with an amplitude of 2 cm . When the particle is at 1 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. `1/(2pisqrt3)`
B. `2pisqrt3`
C. `(2pi)/(sqrt3)`
D. `(sqrt3)/(2pi)`

Answer

To solve the problem step by step, we will use the properties of simple harmonic motion (SHM) and the relationships between velocity, acceleration, and angular frequency.

### Step 1: Understand the given parameters
The amplitude \( A \) of the SHM is given as \( 2 \, \text{cm} \) (which is \( 0.02 \, \text{m} \)), and the position \( x \) of the particle from the mean position is \( 1 \, \text{cm} \) (which is \( 0.01 \, \text{m} \)).

### Step 2: Write the equations for velocity and acceleration
In SHM:
- The magnitude of velocity \( v \) at position \( x \) is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
- The magnitude of acceleration \( a \) at position \( x \) is given by:
\[ a = -\omega^2 x \]
(We will consider the magnitude, so we will take \( a = \omega^2 x \)).

### Step 3: Set the magnitudes of velocity and acceleration equal
According to the problem, at \( x = 1 \, \text{cm} \), the magnitudes of velocity and acceleration are equal:
\[ \omega \sqrt{A^2 - x^2} = \omega^2 x \]
We can cancel \( \omega \) from both sides (assuming \( \omega \neq 0 \)):
\[ \sqrt{A^2 - x^2} = \omega x \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ A^2 - x^2 = \omega^2 x^2 \]
Rearranging this, we get:
\[ A^2 = \omega^2 x^2 + x^2 \]
\[ A^2 = x^2 (\omega^2 + 1) \]

### Step 5: Solve for \( \omega^2 \)
From the equation \( A^2 = x^2 (\omega^2 + 1) \), we can express \( \omega^2 \):
\[ \omega^2 = \frac{A^2}{x^2} - 1 \]

### Step 6: Substitute the values of \( A \) and \( x \)
Substituting \( A = 2 \, \text{cm} \) and \( x = 1 \, \text{cm} \):
\[ \omega^2 = \frac{(2)^2}{(1)^2} - 1 = 4 - 1 = 3 \]
Thus, we have:
\[ \omega = \sqrt{3} \, \text{rad/s} \]

### Step 7: 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 for \( \omega \):
\[ T = \frac{2\pi}{\sqrt{3}} \]

### Final Answer
Thus, the time period \( T \) in seconds is:
\[ T = \frac{2\pi}{\sqrt{3}} \, \text{s} \]

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