A particle oscillating in simple harmonic motion has amplitude 'a'. The distance from the mean position at which its velocity will be one half of the maximum velocity is

Options

A. `(sqrt(3))/(2) a `
B. `sqrt(3) a`
C. `(2)/(sqrt(3))`a
D. `(1)/(sqrt(3))` a

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the problem
We need to find the distance \( x \) from the mean position where the velocity of a particle in simple harmonic motion (SHM) is half of its maximum velocity.

### Step 2: Write down the formulas
The velocity \( V \) of a particle in SHM at a distance \( x \) from the mean position is given by:
\[ V = \omega \sqrt{a^2 - x^2} \]
where:
- \( V \) is the velocity at position \( x \)
- \( \omega \) is the angular frequency
- \( a \) is the amplitude

The maximum velocity \( V_{max} \) occurs at the mean position (when \( x = 0 \)):
\[ V_{max} = a \omega \]

### Step 3: Set up the equation for half of the maximum velocity
According to the problem, we need to find \( x \) when the velocity \( V \) is half of the maximum velocity:
\[ V = \frac{1}{2} V_{max} = \frac{1}{2} (a \omega) = \frac{a \omega}{2} \]

### Step 4: Substitute into the velocity equation
Now, substitute \( V = \frac{a \omega}{2} \) into the velocity equation:
\[ \frac{a \omega}{2} = \omega \sqrt{a^2 - x^2} \]

### Step 5: Simplify the equation
We can divide both sides by \( \omega \) (assuming \( \omega \neq 0 \)):
\[ \frac{a}{2} = \sqrt{a^2 - x^2} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{a}{2}\right)^2 = a^2 - x^2 \]
\[ \frac{a^2}{4} = a^2 - x^2 \]

### Step 7: Rearrange the equation
Rearranging the equation, we get:
\[ x^2 = a^2 - \frac{a^2}{4} \]
\[ x^2 = a^2 \left(1 - \frac{1}{4}\right) \]
\[ x^2 = a^2 \left(\frac{3}{4}\right) \]

### Step 8: Solve for \( x \)
Taking the square root of both sides gives:
\[ x = a \sqrt{\frac{3}{4}} = \frac{a \sqrt{3}}{2} \]

### Final Answer
The distance from the mean position at which the velocity will be one half of the maximum velocity is:
\[ x = \frac{a \sqrt{3}}{2} \]

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