A particle starts simple harmonic motion from the mean position. Its amplitude is a and time period is T. what is its displacement when its speed is half of its maximum speed?

Options

A. `(sqrt(2))/(3)a`
B. `(sqrt(3))/(2)a`
C. `(2)/(sqrt(3))a`
D. `(a)/(sqrt(2))`

Answer

To solve the problem, we need to find the displacement \( x \) of a particle performing simple harmonic motion (SHM) when its speed is half of its maximum speed. Here's a step-by-step solution:

### Step 1: Understand the maximum speed in SHM
The maximum speed \( v_{\text{max}} \) of a particle in SHM is given by the formula:
\[ v_{\text{max}} = A \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

### Step 2: Determine the speed when it is half of maximum speed
Since we need to find the displacement when the speed \( v \) is half of the maximum speed, we can write:
\[ v = \frac{1}{2} v_{\text{max}} = \frac{1}{2} A \omega \]

### Step 3: Use the formula for speed in SHM
The speed \( v \) of a particle at a displacement \( x \) from the mean position is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
Now, we can substitute \( v \) from Step 2 into this equation:
\[ \frac{1}{2} A \omega = \omega \sqrt{A^2 - x^2} \]

### Step 4: Simplify the equation
We can cancel \( \omega \) from both sides (assuming \( \omega \neq 0 \)):
\[ \frac{1}{2} A = \sqrt{A^2 - x^2} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{1}{2} A\right)^2 = A^2 - x^2 \]
This simplifies to:
\[ \frac{1}{4} A^2 = A^2 - x^2 \]

### Step 6: Rearrange the equation to solve for \( x^2 \)
Rearranging the equation, we get:
\[ x^2 = A^2 - \frac{1}{4} A^2 \]
\[ x^2 = \frac{3}{4} A^2 \]

### Step 7: Take the square root to find \( x \)
Taking the square root of both sides gives:
\[ x = \sqrt{\frac{3}{4} A^2} = \frac{\sqrt{3}}{2} A \]

### Final Answer
Thus, the displacement \( x \) when the speed is half of the maximum speed is:
\[ x = \frac{\sqrt{3}}{2} A \]