The time period of oscillation of a `SHO` is `(pi)/(2)s`. Its acceleration at a phase angle `(pi)/(3) rad` from exterme position is `2ms^(-1)`. What is its velocity at a displacement equal to half of its amplitude form mean position? (in `ms^(-1)`

Options

A. `0.707`
B. `0.866`
C. `sqrt(2)`
D. `sqrt(3)`

Explanation

To solve the problem, we need to find the velocity of a simple harmonic oscillator (SHO) at a displacement equal to half of its amplitude from the mean position. Let's break down the solution step by step.

### Step 1: Determine the angular frequency (ω)
The time period \( T \) of the oscillator is given as \( \frac{\pi}{2} \) seconds. We can find the angular frequency \( \omega \) using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{\frac{\pi}{2}} = 4 \, \text{rad/s} \]

### Step 2: Understanding the acceleration at a phase angle
The acceleration \( a \) at a phase angle \( \phi \) in SHM is given by:
\[ a = -\omega^2 A \cos(\phi) \]
We know that the acceleration at a phase angle of \( \frac{\pi}{3} \) radians from the extreme position is \( 2 \, \text{m/s}^2 \). Since the phase angle from the extreme position is \( \frac{\pi}{3} \), we can substitute into the equation:
\[ 2 = -\omega^2 A \cos\left(\frac{\pi}{3}\right) \]
Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ 2 = -\omega^2 A \cdot \frac{1}{2} \]
This simplifies to:
\[ 2 = -\frac{1}{2} \cdot (4^2) A \]
\[ 2 = -8A \]
Thus:
\[ A = -\frac{2}{8} = -\frac{1}{4} \, \text{m} \]
Taking the magnitude, we have \( A = \frac{1}{4} \, \text{m} \).

### Step 3: Calculate the velocity at half the amplitude
The velocity \( v \) in SHM at any displacement \( x \) can be calculated using the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
We need to find the velocity when the displacement \( x \) is half of the amplitude:
\[ x = \frac{A}{2} = \frac{1/4}{2} = \frac{1}{8} \, \text{m} \]
Now substituting \( A \) and \( x \) into the velocity formula:
\[ v = 4 \sqrt{\left(\frac{1}{4}\right)^2 - \left(\frac{1}{8}\right)^2} \]
Calculating the squares:
\[ v = 4 \sqrt{\frac{1}{16} - \frac{1}{64}} \]
Finding a common denominator:
\[ \frac{1}{16} = \frac{4}{64} \]
So:
\[ v = 4 \sqrt{\frac{4}{64} - \frac{1}{64}} = 4 \sqrt{\frac{3}{64}} = 4 \cdot \frac{\sqrt{3}}{8} = \frac{\sqrt{3}}{2} \, \text{m/s} \]
Calculating \( \frac{\sqrt{3}}{2} \):
\[ v \approx 0.866 \, \text{m/s} \]

### Final Answer
The velocity at a displacement equal to half of its amplitude from the mean position is approximately \( 0.866 \, \text{m/s} \).

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