Step 1: Understanding the Concept:
In Simple Harmonic Motion (S.H.M.), the velocity of a particle depends on its displacement from the mean position.
The speed is maximum at the mean position and zero at the extreme positions.
We are given a condition where the instantaneous speed is half of the maximum speed, and we need to find the corresponding displacement.
Step 2: Key Formula or Approach:
The velocity \( v \) of a particle in S.H.M. at a displacement \( x \) is given by:
\[ v = \omega \sqrt{a^2 - x^2} \]
where \( \omega \) is the angular frequency and \( a \) is the amplitude.
The maximum speed \( V_{\text{max}} \) occurs at the mean position (\( x=0 \)) and is given by:
\[ V_{\text{max}} = \omega a \]
Step 3: Detailed Explanation:
According to the given condition, the speed \( u \) at a certain instant is half of the maximum speed:
\[ u = \frac{1}{2} V_{\text{max}} \]
Substitute the expressions for \( u \) (which is \( v \) at displacement \( x \)) and \( V_{\text{max}} \) into this equation:
\[ \omega \sqrt{a^2 - x^2} = \frac{1}{2} (\omega a) \]
Since \( \omega \) is non-zero, we can cancel it from both sides:
\[ \sqrt{a^2 - x^2} = \frac{a}{2} \]
To eliminate the square root, square both sides of the equation:
\[ a^2 - x^2 = \left(\frac{a}{2}\right)^2 \]
\[ a^2 - x^2 = \frac{a^2}{4} \]
Now, solve for \( x^2 \):
\[ x^2 = a^2 - \frac{a^2}{4} \]
\[ x^2 = \frac{3a^2}{4} \]
Taking the square root of both sides to find the displacement \( x \):
\[ x = \pm \frac{\sqrt{3}a}{2} \]
The magnitude of the displacement is \( \frac{\sqrt{3}a}{2} \).
Step 4: Final Answer:
The displacement of the particle at that instant is \( \frac{\sqrt{3}a}{2} \).