Question

The position of a 0.30-kg object attached to a spring is described by

x = (0.22 m) cos(0.3?t)

(a) Find the amplitude of the motion.
m

(b) Find the spring constant.
N/m

(c) Find the position of the object at t = 0.26 s.
m

(d) Find the object's speed at t = 0.26 s.
m/s

Answer 1

(a)

Amplitude is, 0.22 m

(b)

The spring constant is,

        k = w²m

          = [(0.3)(pi)]²(0.30 kg)

          = 0.266 N/m

(c)

The position of the object is defined as,

      x = (0.22 m) cos(0.3(pi)t).

At t = 0.26 s, the position is,

      x = (0.22 m) cos(0.3(pi)(0.26) s)

         = 0.213 m

(b)

The speed of the object is,

      dx/dt = d/dt [(0.22 m) cos(0.3(pi)t)]

               = (0.22 m) [-(0.3)(pi)]sin(0.3(pi)t)

               = (0.22 m) [-(0.3)(pi)]sin(0.3(pi)(0.26 s))

               = -0.0503 m/s