Question

A particle is performing S.H.M whose distance from mean position varies as \(x = Asin(wt)\). Find the position of the particle from the mean position, where kinetic energy and potential energy is equal.

• A. \(\frac{A}{2}\)
• B. \(\frac{A}{\sqrt2}\)
• C. \(\frac{A}{2\sqrt2}\)
• D. \(\frac{A}{4}\)

Answer 1

The Correct Option is B

For a particle in simple harmonic motion (SHM), the total energy \( E \) is the sum of kinetic energy (K.E.) and potential energy (P.E.), and is given by: \[ E = \frac{1}{2}M \omega^2 A^2, \] where \( M \) is the mass, \( \omega \) is the angular frequency, and \( A \) is the amplitude. When the kinetic energy equals the potential energy: \[ \text{K.E.} = \text{P.E.}. \] At any point in SHM, the kinetic energy is: \[ \text{K.E.} = \frac{1}{2}M \omega^2 (A^2 - x^2), \] and the potential energy is: \[ \text{P.E.} = \frac{1}{2}M \omega^2 x^2. \] Setting \( \text{K.E.} = \text{P.E.} \): \[ \frac{1}{2}M \omega^2 (A^2 - x^2) = \frac{1}{2}M \omega^2 x^2. \] Simplify: \[ A^2 - x^2 = x^2. \] Rearrange: \[ 2x^2 = A^2. \] Solve for \( x \): \[ x^2 = \frac{A^2}{2}. \] \[ x = \pm \frac{A}{\sqrt{2}}. \] Thus, the distance from the mean position when the kinetic energy equals the potential energy is \( \boxed{\frac{1}{\sqrt{2}}A} \).