If the maximum velocity of a particle in SHM is `v_0` then its velocity at half the amplitude from position of rest will be :

Options

A. `v_0//2`
B. `v_0`
C. ` v_0 sqrt(3//2)`
D. `v_0 sqrt3//2`

Answer

To solve the problem, we need to find the velocity of a particle in Simple Harmonic Motion (SHM) when it is at half the amplitude from the position of rest. Let's break this down step by step.

### Step 1: Understand the formula for velocity in SHM
In SHM, the velocity \( v \) of a particle at a displacement \( x \) from the mean position is given by the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
where:
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude,
- \( x \) is the displacement from the mean position.

### Step 2: Identify the displacement
Since we need to find the velocity at half the amplitude, we have:
\[ x = \frac{A}{2} \]

### Step 3: Substitute the displacement into the velocity formula
Now, substituting \( x = \frac{A}{2} \) into the velocity formula:
\[ v = \omega \sqrt{A^2 - \left(\frac{A}{2}\right)^2} \]

### Step 4: Simplify the expression
Calculating \( \left(\frac{A}{2}\right)^2 \):
\[ \left(\frac{A}{2}\right)^2 = \frac{A^2}{4} \]
Now substitute this back into the equation:
\[ v = \omega \sqrt{A^2 - \frac{A^2}{4}} = \omega \sqrt{\frac{4A^2}{4} - \frac{A^2}{4}} = \omega \sqrt{\frac{3A^2}{4}} = \omega \cdot \frac{A\sqrt{3}}{2} \]

### Step 5: Relate maximum velocity to angular frequency and amplitude
The maximum velocity \( v_0 \) in SHM is given by:
\[ v_{max} = \omega A \]
From this, we can express \( \omega \) in terms of \( v_0 \) and \( A \):
\[ \omega = \frac{v_0}{A} \]

### Step 6: Substitute \( \omega \) back into the velocity equation
Now substitute \( \omega \) into the velocity equation we derived:
\[ v = \left(\frac{v_0}{A}\right) \cdot \frac{A\sqrt{3}}{2} = \frac{v_0 \sqrt{3}}{2} \]

### Final Answer
Thus, the velocity of the particle at half the amplitude from the position of rest is:
\[ v = \frac{v_0 \sqrt{3}}{2} \]

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