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) = (0.26 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

Please be detailed.

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 is A = 0.26m.

B) w = sqrt(k/m)

spring constant k = m*w^2

                          = 0.56*6^2

                          = 20.16 N/m

C) x = A*cos(wt)
when t = T/2
and w = 2*pi/T

x = A*cos(w*T/2)

   = A*cos(pi)

   = -A

= -0.26m

D) when t = T/3...
x= A*cos(wt) = A*cos(2*pi/3)

    = -0.5*A

    = -0.13m

e) when x = -0.1m

-0.1 = 0.26*cos(6t)

cos(6t) = -0.384

6t = acos(-0.384)

    = 1.965
t = 1.965/6

= 0.3275 sec

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