Question

The amplitude of a simple harmonic oscillator is A. When the velocity of particle is half of its maximum velocity, then its position is at

• A. \(\frac{A}{2}\)
• B. \(\frac{\sqrt{3}A}{4}\)
• C. \(\frac{A}{4}\)
• D. \(\frac{\sqrt{3}A}{2}\)

Hint:

The key relationship \(v = \omega \sqrt{A^2 - x^2}\) is fundamental to SHM. It's derived from the conservation of energy in the oscillator system. Memorizing this formula is essential for solving problems that relate position and velocity in SHM.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the position (\(x\)) of a particle in Simple Harmonic Motion (SHM) when its velocity (\(v\)) is half of its maximum possible velocity (\(v_{max}\)).

Step 2: Key Formula or Approach:
The velocity of a particle in SHM as a function of its position \(x\) is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
where \(\omega\) is the angular frequency and \(A\) is the amplitude.
The maximum velocity occurs at the equilibrium position (\(x=0\)) and is given by:
\[ v_{max} = A\omega \]

Step 3: Detailed Explanation:
We are given the condition that \(v = \frac{1}{2}v_{max}\).
Substitute the formulas for \(v\) and \(v_{max}\) into this condition:
\[ \omega \sqrt{A^2 - x^2} = \frac{1}{2}(A\omega) \]
The angular frequency \(\omega\) cancels from both sides:
\[ \sqrt{A^2 - x^2} = \frac{A}{2} \]
To solve for \(x\), square both sides of the equation:
\[ A^2 - x^2 = \left(\frac{A}{2}\right)^2 = \frac{A^2}{4} \]
Now, isolate \(x^2\):
\[ x^2 = A^2 - \frac{A^2}{4} = \frac{4A^2 - A^2}{4} = \frac{3A^2}{4} \]
Take the square root of both sides to find the position \(x\):
\[ x = \pm \sqrt{\frac{3A^2}{4}} = \pm \frac{\sqrt{3}A}{2} \]
The question asks for the position, and the positive value is given in the options.

Step 4: Final Answer:
The position of the particle is at \(\frac{\sqrt{3}A}{2}\).