Question 68: Oscillations

Question

A particle executes SHM of amplitude A. The distance from the mean position when its's kinetic energy becomes equal to its potential energy is :

Accepted Answer

The total energy of a particle executing simple harmonic motion (SHM) is given by:
$$E = \frac{1}{2}m\omega^2A^2$$
where $$m$$ is the mass of the particle, $$\omega$$ is the angular frequency of the SHM, and $$A$$ is the amplitude of the motion.
At any point during SHM, the kinetic energy of the particle is given by:
$$K = \frac{1}{2}m\omega^2(x^2 + A^2\cos^2\omega t)$$
where $$x$$ is the displacement of the particle from the mean position.
The potential energy of the particle at the same point is given by:
$$U = \frac{1}{2}m\omega^2(x^2 + A^2\sin^2\omega t)$$
When the kinetic energy becomes equal to the potential energy, we have:
$$K = U$$
$$\frac{1}{2}m\omega^2(x^2 + A^2\cos^2\omega t) = \frac{1}{2}m\omega^2(x^2 + A^2\sin^2\omega t)$$
Simplifying this equation, we get:
$$x^2 = \frac{1}{2}A^2$$
$$x = \pm\frac{1}{\sqrt{2}}A$$
Therefore, the distance from the mean position when the kinetic energy becomes equal to the potential energy is $$\boxed{\frac{1}{\sqrt{2}}A}$$

Answer

$$\frac{1}{\sqrt{2}} A$$

Answer

$$\frac{1}{2} A$$

Answer

$$2 \mathrm{~A}$$

Answer

$$\sqrt{2 A}$$

Oscillations

A particle executes SHM of amplitude A. The distance from the mean position when its's kinetic energy becomes equal to its potential energy is :

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