A mass of 0.34 kg is attached to a spring and set into oscillation on a horizont

Question

A mass of 0.34 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) = (0.48 m)cos[(20 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)  amplitude of oscillation for the oscillating mass
0.48 m

(b) force constant for the spring
For the spring-mass system,
² = k/m
(20 rad/s)² = k / (0.34 kg)
k = 136 N/m

(c)  position of the mass after it has been oscillating for one half a period
x(0.5 s) = (0.48 m)cos[(20 rad/s)(0.5 s)]
x(0.5 s) = 0.38 m

(d) position of the mass one-third of a period after it has been released
x(t) = (0.48 m)cos[2/3]
x(t) = 0.47 m

(e)  time it takes the mass to get to the position x = 0.10 m after it has been released
-0.10 m = (0.48 m)cos[(20 rad/s)(t)]
cos(20t) = -1/0.48
20t = 0.20
t = 0.11 s

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