A mass of 0.16 kg is attached to a spring and set into oscillation on a horizont

Question

A mass of 0.16 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) = (0.50 m)cos[(18 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

x(t) = (0.50 m)cos[(18 rad/s)t].
somparing it with general oscillation equation:
x(t)= A* cos (w*t)
we get,
A =0.5 m
w = 18 rad/s

a)
Ampitude = 0.5 m

b)
use:
w = sqrt(K/m)
18 =sqrt (K/0.16)
K=51.84 N/m

c)
period, T = 2*pi/w= 2*pi/18 =0.35 s
so,
t = T/2
= 0.35/2
=0.175 s

x(t) = (0.50 m)cos[(18 rad/s)t].
x(0.175)= (0.50 m)cos[(18 rad/s)0.175 s].
= 0.5*(-1)
= -0.5 m

d)
t = T/3 = 0.35/3=0.117 S
x(t) = (0.50 m)cos[(18 rad/s)t].
x(0.117)= (0.50 m)cos[(18 rad/s)0.117 s].
= 0.5 *(-0.51)
= -0.255 M

Only 4 parts at a time

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