A mass of 0.56 kg is attached to a spring and set into oscillation on a horizont

Question

A mass of 0.56 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) = (0.46 m)cos[(12 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) x(t) =(0.46m)cos(12t)

A =0.46 m

(b)

For the spring-mass system,

² = k/m

(12 rad/s)² = k / (0.56 kg)
k = 80.64 N/m

(c)
x(T/2) = (0.46 m)cos[(2pi/T)*(T/2)]

x= - 0.46 m

(d)
x(t) = (0.46 m)cos[(2pi/T)*(T/3)]

x = -0.23 m

(e)
-0.10 m = (0.46 m)cos[(12 rad/s)(t)]

cos(12t) = -0.2174

t = 0.15 s

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