Amplitude of a harmonic oscillator is A ,when velocity of particles is half of maximum velocity ,then determine the position of particle.

Options

A. `sqrt(v_(1)v_(2))/g`
B. `(v_(1)sqrt(v_(1)+v_(2)))/g`
C. `((v_(1)+ v_(2))sqrt(v_(1)v_(2)))/g`
D. `(v_(2)sqrt(v_(1)v_(2)))/g`

Answer

To solve the problem step by step, we will use the concepts of harmonic motion and the relationship between velocity, amplitude, and position.

### Step 1: Understand the given parameters
We are given:
- Amplitude of the harmonic oscillator, \( A \)
- Velocity of the particle, \( V \), which is half of the maximum velocity, \( V_{max} \).

### Step 2: Write the expression for maximum velocity
The maximum velocity \( V_{max} \) of a harmonic oscillator is given by:
\[ V_{max} = A \omega \]
where \( \omega \) is the angular frequency.

### Step 3: Express the velocity when it is half of maximum velocity
Since the velocity \( V \) is half of the maximum velocity, we can write:
\[ V = \frac{1}{2} V_{max} = \frac{1}{2} A \omega \]

### Step 4: Use the formula for velocity in harmonic motion
The velocity \( V \) of a particle in harmonic motion is also expressed as:
\[ V = \omega \sqrt{A^2 - x^2} \]
where \( x \) is the position of the particle.

### Step 5: Set the two expressions for velocity equal
Now, we can set the two expressions for \( V \) equal to each other:
\[ \frac{1}{2} A \omega = \omega \sqrt{A^2 - x^2} \]

### Step 6: Cancel \( \omega \) from both sides
Assuming \( \omega \neq 0 \), we can divide both sides by \( \omega \):
\[ \frac{1}{2} A = \sqrt{A^2 - x^2} \]

### Step 7: 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 8: Rearrange the equation to solve for \( x^2 \)
Rearranging the equation:
\[ x^2 = A^2 - \frac{1}{4} A^2 \]
\[ x^2 = \frac{4}{4} A^2 - \frac{1}{4} A^2 \]
\[ x^2 = \frac{3}{4} A^2 \]

### Step 9: Solve for \( x \)
Taking the square root of both sides gives:
\[ x = \pm \sqrt{\frac{3}{4}} A = \pm \frac{\sqrt{3}}{2} A \]

### Final Answer
Thus, the position of the particle when its velocity is half of the maximum velocity is:
\[ x = \pm \frac{\sqrt{3}}{2} A \]

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