Question

A mass is attatched to a spring, displaced and then released from rest with an angular velocity of 0.5 rad/s. Determine the time hen the kinetic energy and potential energy are first equal

Answer 1

Answer on Question #65192, Physics Mechanics Relativity

A mass is attached to a spring, displaced and then released from rest with an angular velocity of 0.5 rad/s. Determine the time when the kinetic energy and potential energy are first equal.

Solution:

In this situation, we have harmonic oscillations: $x(t) = x_{\text{max}} \cos(\omega t)$ ; (t=0; x=A-amplitude)

$$\omega - \text{angular velocity} = \frac{0.5 \text{rad}}{s}$$

When kinetic energy and potential energy are equal: $E_{mech} = E_{pot1} + E_{kin1} = 2E_{pot1} = 2E_{kin1}$ ; $E_{pot}(t) = \frac{\pi * x(t)^2}{2}$ ; According to law of conservation of energy:

$$E_{mech} = \text{const}; E_{mech} = 2E_{pot1} = E_{pot(\text{max})} \Rightarrow 2 * \frac{\pi * x_1^2}{2} = \frac{\pi * x_{\text{max}}^2}{2} \Rightarrow 2 * x_1^2 = x_{\text{max}}^2;$$ $$2 * (x_{\text{max}} \cos(\omega t))^2 = x_{\text{max}} \Rightarrow \cos(\omega t) = \frac{1}{\sqrt{2}}; \Rightarrow \omega t = \frac{\pi}{4}; t = \frac{\pi}{4\omega} = \frac{3.14}{4 + 0.5} = 1.57s$$

Answer: $t = 1.57s$

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