Question

The kinetic energy of a particle of mass m executing linear simple harmonic motion with angular velocity $ω$ and amplitude $a$ is $\frac 1{4} ma^2ω^2$ at a distance of _______from the mean position.

• A. $\frac{a}{8} $
• B. $\frac{a}{2} $
• C. $\frac{a}{4} $
• D. $ $
• E. $\frac{a}{\sqrt2} $

Answer 1

Answer: (E)

The kinetic energy \( K \) of a particle of mass \( m \) executing simple harmonic motion (SHM) with angular velocity \( \omega \) and amplitude \( a \) is given by the equation:

\[ K = \frac{1}{2} m v^2 \] Where \( v \) is the velocity of the particle at a given position. For a particle in SHM, the velocity at a distance \( x \) from the mean position is: \[ v = \omega \sqrt{a^2 - x^2} \] The kinetic energy at distance \( x \) from the mean position can therefore be written as: \[ K = \frac{1}{2} m \left( \omega \sqrt{a^2 - x^2} \right)^2 = \frac{1}{2} m \omega^2 (a^2 - x^2) \] We are given that the kinetic energy is \( \frac{1}{4} ma^2 \omega^2 \). Setting this equal to the expression for kinetic energy, we get: \[ \frac{1}{4} ma^2 \omega^2 = \frac{1}{2} m \omega^2 (a^2 - x^2) \] Simplifying the equation: \[ \frac{1}{4} a^2 = \frac{1}{2} (a^2 - x^2) \] Multiplying both sides by 2: \[ \frac{1}{2} a^2 = a^2 - x^2 \] Solving for \( x^2 \): \[ x^2 = \frac{a^2}{2} \] Taking the square root: \[ x = \frac{a}{\sqrt{2}} \] Therefore, the particle is at a distance of \( \frac{a}{\sqrt{2}} \) from the mean position when its kinetic energy is \( \frac{1}{4} ma^2 \omega^2 \).

Correct Answer:

Correct Answer: (E) \( \frac{a}{\sqrt{2}} \)