A mass of 0.12 kg is attached to a spring and set into oscillation on a horizont

Question

A mass of 0.12 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.44 m)cos[(6 rad/s)t]. Determine the following. (a) amplitude of oscillation for the oscillating mass m (b) force constant for the spring N/m (c) position of the mass after it has been oscillating for one half a period m (d) position of the mass one-third of a period after it has been released m (e) time it takes the mass to get to the position x = 0.10 m after it has been released s

Explanation / Answer

In general, x(t) = A cos(t - ), where A is the amplitude, is the angular frequency, and is some phase shift.

(a) 0.44 m .....Ans.

(b) For the spring-mass system,
² = k/m
(6 rad/s)² = k / (0.12 kg)
k = 4.32 N/m .......Ans.

(c) x(0.5 s) = (0.44 m)cos[(6 rad/s)(0.5 s)]
x(0.5 s) = -0.4355 m

(d) x(t) = (0.44 m)cos[2/3]
x(t) = -0.358 m

(e)
-0.10 m = (0.44 m)cos[(6 rad/s)(t)]
cos(6t) = -1/4.4
6t = 1.3415
t = 0.2235 s

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