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I have been attempting to complete this problm for many days now, but i keep getting stuck. I have no idea what to possibly do. Any help would be appreciated.

A 200 g mass attached to a horizontal spring oscillates at a frequency of 2.0 Hz. At t = 0 s, the mass
is at x = 5.0 cm and has vx = -30 cm/s. Determine
a) The period,
b) The angular frequency,
c) The amplitude,
d) The phase constant,
e) The maximum acceleration,
f) The position at t = 0.4 s, and
g) Write down an equation that describes the position of the oscillating mass as a function of time ( ie. what is x(t)? ).

Answer
Hello Nick,

Two things that your problem doesn't explain, but I have to assume:
1) the x measurement is such that x = 0 is the  position in the center of its oscillation. The position at which it would remain at rest if you stopped it there.
2) the system oscillates without friction so that energy is conserved.  

a) The period T = 1/f = 1/2 Hz = 0.5 seconds

b) The angular frequency, w = 2*pi*f = 2*pi*2 Hz =  12.6 radians/s

c) One of the results of the analysis of the simple harmonic motion equation in my book and I hope in your book is that
w^2 = k/m
(12.6 radians/s)^2 = k/0.2 kg
k = 0.2 kg*(12.6 radians/s)^2 = 31.6 kg/s^2
We want the spring constant to be in units of N/m, so multiplying and dividing by m,
k = 31.6 kg.m/m.s^2 = 31.6 N/m

Due to conservation of energy, the sum of the spring potential energy and kinetic energy at time t=0 is equal to the Total system energy, TSE, at any time. So we can get A because when the mass is at the extreme position, velocity is zero and all system energy is in the form of spring potential energy.

TSE = SPE(t=0) + KE(t=0) = (1/2)*k*x^2 + (1/2)*m*v^2
TSE = (1/2)*(31.6 N/m)*(0.05 m)^2 + (1/2)*0.2 kg*(-0.3 m/s)^2 = 0.0395 J + 0.009 J
TSE = 0.0485 J

When at the extreme position, x = A,
TSE = SPE(x=A) = (1/2)*k*x^2
TSE = 0.0485 J = (1/2)*(31.6 N/m)*x^2 (Note: 1 J = 1 N.m)
x = sqrt(2*0.0485 N.m/(31.6 N/m) = sqrt(0.00153 m^2) = 0.0554 m
So A = 0.0554 m

d) The system is described by the standard simple harmonic motion equation
x = A*cos(w*t + delta)

We know that w = 12.6 radians/s, A = 0.0554 m, and when t = 0, x = 0.050 m. Plugging that into the standard simple harmonic motion equation
x = 0.050 m = 0.0554 m*cos(0 + delta)
cos(delta) = 0.050/0.0554 = 0.902
delta = arccos0.902 = 25.5 degrees

e) The maximum acceleration will occur when x = A and the mass has zero velocity but is reversing direction.
Since x = A*cos(w*t + delta) and cos(0) = 1, w*t + delta must be zero.
The first derivative, dx/dt, of x will yield an expression for velocity as a function of time.
x = A*cos(w*t + delta)
dx/dt = -w*A*sin(w*t + delta)
The second derivative, d2x/dt2, will yield an expression for acceleration. Do that and plug in the data for x = A.

f) x = A*cos(w*t + delta) Plug in t = 0.4 s and the other data.

e) x = A*cos(w*t + delta)
Like I said above, I hope your book gives you that equation, mine does. This is the solution of the differential equation
m*d2x/dt2 + k*x = 0

Please verify my math.

I hope this helps,
Steve

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