Question

A particle is vibrating in 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. 10 cm
• B. $\sqrt{2}$cm
• C. 2 cm
• D. $2\sqrt{2}$cm

Answer 1

Answer: (D)

$2\sqrt{2}$cm

Answer 2

The total energy E of a particle vibrating SHM is given by

E = $\frac{1}{2}$mω²a² .....(1)

The kinetic energy K is given by

K =$\frac{1}{2}$mω² (a² – y²)

where y = displacements of the particle

but K = $\frac{E}{2} = \frac{1}{2}\left\lbrack \frac{1}{2}m\omega^{2}a^{2} \right\rbrack$

∴ $\frac{1}{2}\left\lbrack \frac{1}{2}m\omega^{2}a^{2} \right\rbrack = \frac{1}{2}m\omega^{2}(a^{2} - y^{2})$ or $\frac{a^{2}}{2} = a^{2} - y^{2}$ or

y² = $\frac{a^{2}}{2}$ ∴ y = $\frac{a}{\sqrt{2}}$

Hence the kinetic energy is half of the total energy when displacement of the particle is a/$\sqrt{2}$. Given that a = 4cm.

∴ y = 4/$\sqrt{2}$ = 2$\sqrt{2}$.