Question

The displacement of a block attached to a horizontal spring whose spring constant is 6 N/m is given by

x = 0.4 \cos(4.1 t - 0.8) m.

What is the earliest time (t > 0) when the kinetic energy equals one-half the potential energy?

Answer 1

Answer on Question 57618, Physics – Mechanics | Relativity

Question:

The displacement of a block attached to a horizontal spring whose spring constant is 6 N/m is given by

$$x = 0.4 \cdot \cos(4.1 t - 0.8) \, \text{m}.$$

What is the earliest time (t > 0) when the kinetic energy equals one-half the potential energy?

Answer:

$$PE = 2 \, KE$$ $$k \frac{x^2}{2} = 2 \, m \frac{\dot{x}^2}{2}$$ $$\dot{x} = -1.64 \cos(4.1 t - 0.8)$$ $$\omega^2 x^2 = 2 \dot{x}^2$$ $$\omega^2 x^2 = 2 \dot{x}^2, \quad \text{where} \quad \omega = 4.1 \frac{\text{rad}}{\text{s}}$$ $$\omega^2 \cdot 0.16 \cdot \cos^2(4.1 t - 0.8) = 2 \cdot 2.7 \cdot \sin^2(4.1 t - 0.8)$$ $$\tan(4.1 t - 0.8) = 0.707$$ $$4.1 t - 0.8 = 0.614$$ $$t = 0.34 \, \text{s}$$

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