Question

A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position, is its energy half potential and half kinetic

• A. 1 cm
• B. $\sqrt{2}$cm
• C. 3 cm
• D. $2\sqrt{2}$cm

Answer 1

Answer: (D)

$2\sqrt{2}$cm

Answer 2

Let x be the point where K.E. = P.E.

Hence$\frac{1}{2}m\omega^{2}(a^{2} - x^{2}) = \frac{1}{2}m\omega^{2}x^{2}$

⇒ $2x^{2} = a^{2}$ ⇒ $T_{1} = 2s$