Answer to Question #98571 in Physics for rose

Question

A glider is undergoing simple harmonic motion on a frictionless air track. When it is at position x = 0.30 m, its kinetic energy is half of its potential energy, K=1/2U. What is the amplitude of its simple harmonic motion?

Answer 1

The potential energy of the harmonic oscillator is $U = \frac{m \omega^2 x^2}{2}$, and total energy is $E = T + U = \frac{m \omega^2 A^2}{2}$, where $A$ is the amplitude, $m$ is the mass, $\omega$ is the angular frequency. .

At $x = 0.3 m$, the kinetic energy is $T = \frac{U}{2}$, so at that moment $E = \frac{U}{2} + U = \frac{3 U}{2} = \frac{3}{2} \frac{m \omega^2 x^2}{2}$, and according to the previous equation $E = \frac{m \omega^2 A^2}{2}$. Hence, at $x = 0.3$, $\frac{3}{2} \frac{m \omega^2 x^2}{2} = \frac{m \omega^2 A^2}{2}$, from where $A^2 = \frac{3 x^2}{2}$ and $A = \sqrt{\frac{3}{2}}x \approx 0.37 m$.