An object with a mass of 0.527kg is attached to a long horizontal spring with a

Question

An object with a mass of 0.527kg is attached to a long horizontal spring with a constant of 52.6kg/s2. For a moment when t = 0, the object has a maximum speed of 21.3m/s. (a) What is the amplitude of oscillation? (b) For the center defined as x = 0, where in the motion is the potential energy three times the kinetic energy (take the positive x value for the input)? (c) How long does it take the the particle to move from -0.500m to 0.500m? (d) At small angles with the same object, what length is required for a pendulum to have the same period?

Explanation / Answer

a) 0.5 *0. 527 * 21.3*21.3 = 0.5*52.6*A^2 amplitude = 15.46 m b) 0.5* 0.527*V^2 * 3 = 0.5* 52.6 * X^2 V^2 = 33.27 *X^2 0.5* 0.527*V^2 + 0.5* 52.6 * X^2 = 119.547 1.333*0.5*52.6* X^2 = 119.547 X = 1.848m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.