A particle executes `SHM` along a straight line `4cm` long. When the displacement is `1cm` its velocity and acceleration are numerically equal. The time period of `SHM` is

Options

A. `2pi s`
B. `(2pi)/(sqrt(3))s`
C. `(2pi)/(sqrt(5))s`
D. `(2pi)/(sqrt(7))s`

Explanation

To find the time period of the simple harmonic motion (SHM) of a particle executing motion along a straight line of 4 cm long, we can follow these steps:

### Step 1: Determine the Amplitude
The total length of the motion is 4 cm, which means the amplitude \( A \) is half of this length:
\[ A = \frac{4 \text{ cm}}{2} = 2 \text{ cm} \]

### Step 2: Use the Given Condition
We are given that when the displacement \( x \) is 1 cm, the magnitude of velocity \( |V| \) and acceleration \( |A| \) are equal. Therefore, we can set up the equation:
\[ |V| = |A| \]

### Step 3: Write the Formulas for Velocity and Acceleration
The formulas for velocity \( V \) and acceleration \( A \) in SHM are:
\[ V = \omega \sqrt{A^2 - x^2} \]
\[ A = -\omega^2 x \]
Since we are interested in magnitudes, we can drop the negative sign for acceleration.

### Step 4: Set the Equations Equal
From the condition \( |V| = |A| \), we have:
\[ \omega \sqrt{A^2 - x^2} = \omega^2 x \]

### Step 5: Cancel \( \omega \) (assuming \( \omega \neq 0 \))
We can divide both sides by \( \omega \):
\[ \sqrt{A^2 - x^2} = \omega x \]

### Step 6: Solve for \( \omega \)
Squaring both sides gives:
\[ A^2 - x^2 = \omega^2 x^2 \]
Rearranging this gives:
\[ A^2 = x^2 + \omega^2 x^2 \]
Factoring out \( x^2 \):
\[ A^2 = x^2(1 + \omega^2) \]
Thus,
\[ \omega^2 = \frac{A^2 - x^2}{x^2} \]

### Step 7: Substitute Values
Substituting \( A = 2 \text{ cm} \) and \( x = 1 \text{ cm} \):
\[ \omega^2 = \frac{2^2 - 1^2}{1^2} = \frac{4 - 1}{1} = 3 \]
So,
\[ \omega = \sqrt{3} \]

### Step 8: Find the Time Period
The time period \( T \) is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting \( \omega = \sqrt{3} \):
\[ T = \frac{2\pi}{\sqrt{3}} \]

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

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