A particle is executing S.H.M. along 4 cm long line with time period `(2pi)/(sqrt2)` sec. If the numerical value of its velocity and acceleration is same, then displacement will be

Options

A. 0.5 cm
B. 1.15 cm
C. `2 pi cm`
D. `(2 pi)/(sqrt3)`

Answer

To solve the problem step by step, we need to find the displacement \( x \) of a particle executing Simple Harmonic Motion (S.H.M.) given that the numerical values of its velocity and acceleration are the same.

### Step 1: Identify the parameters
The length of the line (amplitude) \( A = 4 \, \text{cm} \) and the time period \( T = \frac{2\pi}{\sqrt{2}} \, \text{s} \).

### Step 2: Calculate angular frequency \( \omega \)
The angular frequency \( \omega \) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{\frac{2\pi}{\sqrt{2}}} = \sqrt{2} \, \text{rad/s} \]

### Step 3: Write the equations for velocity and acceleration
The velocity \( v \) in S.H.M. is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
The acceleration \( a \) is given by:
\[ a = -\omega^2 x \]
Since we are interested in the magnitudes, we can write:
\[ |v| = \omega \sqrt{A^2 - x^2} \]
\[ |a| = \omega^2 |x| \]

### Step 4: Set the magnitudes equal
According to the problem, the numerical values of velocity and acceleration are the same:
\[ \omega \sqrt{A^2 - x^2} = \omega^2 |x| \]
We can cancel \( \omega \) (since \( \omega \neq 0 \)):
\[ \sqrt{A^2 - x^2} = \omega |x| \]

### Step 5: Square both sides
Squaring both sides gives:
\[ A^2 - x^2 = \omega^2 x^2 \]
Rearranging this, we get:
\[ A^2 = x^2 + \omega^2 x^2 \]
\[ A^2 = x^2 (1 + \omega^2) \]

### Step 6: Substitute the values
Substituting \( A = 4 \, \text{cm} \) and \( \omega = \sqrt{2} \):
\[ 4^2 = x^2 (1 + (\sqrt{2})^2) \]
\[ 16 = x^2 (1 + 2) \]
\[ 16 = 3x^2 \]

### Step 7: Solve for \( x^2 \)
\[ x^2 = \frac{16}{3} \]
\[ x = \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \approx 2.309 \, \text{cm} \]

### Step 8: Calculate numerical values
The numerical value of \( x \) is approximately \( 2.309 \, \text{cm} \).

### Conclusion
Thus, the displacement \( x \) where the numerical values of velocity and acceleration are the same is approximately \( 2.309 \, \text{cm} \).