Question:

The energy of a particle executing Simple Harmonic Motion (SHM) is given by E = Ax^2 + BV^2. Here 'x' is the displacement of the particle from its mean position, 'V' is its velocity at 'x', and A and B are positive constants. The maximum velocity of the particle is:

Show Hint

In SHM: \[ v_{\max} \text{ occurs at mean position where } x = 0 \]
Updated On: Jun 10, 2026
  • \( \sqrt{\frac{E}{B}} \)
  • \( \sqrt{\frac{E}{A}} \)
  • \( \sqrt{\frac{2E}{B}} \)
  • \( \sqrt{\frac{2E}{A}} \)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

Concept: In Simple Harmonic Motion, total mechanical energy remains conserved and is given by: \[ E = \text{Kinetic Energy} + \text{Potential Energy} \] The given expression: \[ E = Ax^2 + Bv^2 \] shows that kinetic energy part is \(Bv^2\).

Step 1: Maximum velocity condition Velocity is maximum at mean position: \[ x = 0 \] Substitute: \[ E = Bv_{\max}^2 \]

Step 2: Solve for maximum velocity \[ v_{\max}^2 = \frac{E}{B} \] \[ v_{\max} = \sqrt{\frac{E}{B}} \] Thus, maximum velocity is: \[ \boxed{\sqrt{\frac{E}{B}}} \]
Was this answer helpful?
0
0

Top AP EAPCET Physics Questions

View More Questions

Top AP EAPCET Simple Harmonic Motion Questions

View More Questions