An object with a mass of 0.427kg is attached to a long horizontal spring with a

Question

An object with a mass of 0.427kg is attached to a long horizontal spring with a constant of 42.6kg/s2. For a moment when t = 0, the object has a maximum speed of 19.3m/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? (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

.5kx^2=.5mv^2

42.6*x^2=.427*19.3^2

x=1.93 m

(b).5kx^2=.5mv^2*3

.5kx^2+.5mv^2=.5m*19.3^2(total energy is constnt
4*.5kx^2=.5m*19.3^2

x=.965 m

(c)

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