A mass of 0.48 kg is attached to a spring and set into oscillation on a horizont

Question

A mass of 0.48 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) = (0.22 m)cos[(8 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)
0.22 m

(b)
For the spring-mass system,
² = k/m
(8 rad/s)² = k / (0.48 kg)
k = 30.72N/m

(c)
x(0.5 s) = (0.22 m)cos[(8 rad/s)(0.5 s)]
x(0.5 s) = 0.14 m

(d)
x(t) = (0.22 m)cos[2/3]
x(t) = -0.11 m

(e)
-0.10 m = (0.22 m)cos[(8 rad/s)(t)]
cos(8t) = -1/5
8t = 1.77215
t = 0.22 s

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