The ratio of kinetic energy and potential energy of a simple harmonic oscillator, at a displacement equal to half its amplitude is given by A. 1 : 2 B. 1 : 1 C. 2 : 1 D. 3 : 1

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The correct answer is $\boxed{\text{B) 1 : 1}}$.

The kinetic energy of a simple harmonic oscillator is given by:

$$K = \frac{1}{2}m\omega^2x^2$$

where $m$ is the mass of the oscillator, $\omega$ is the angular frequency, and $x$ is the displacement from the equilibrium position.

The potential energy of a simple harmonic oscillator is given by:

$$U = \frac{1}{2}kx^2$$

where $k$ is the spring constant.

At a displacement equal to half its amplitude, the kinetic energy and potential energy of the oscillator are equal. This is because the potential energy is at a maximum when the displacement is zero, and the kinetic energy is at a maximum when the displacement is equal to the amplitude.

Therefore, the ratio of kinetic energy and potential energy of a simple harmonic oscillator, at a displacement equal to half its amplitude is $\boxed{\text{1 : 1}}$.

Option A is incorrect because the kinetic energy is not twice the potential energy.

Option C is incorrect because the potential energy is not twice the kinetic energy.

Option D is incorrect because the kinetic energy is not three times the potential energy.

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