Simple Harmonic Motion position and velocity problem

[madchemist]

"A particle moves along the x-axis. It is initially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration -0.320 m/s^2. Assume it moves with simple harmonic motion for 4.50 s and x=0 is its equilibrium position. Find its position and velocity at the end of this time interval."


x=Amplitude * cos2(pi)ft

Found f using T= radical (x/a) * 2pi = 5.769 so f= 0.1734 Hz

However, even with solving for f I'm still left with two unknowns, i.e. A and t. Please help...

 

[Doc Al]

You are given data for the initial postion, velocity, and acceleration. Set up equations for each. That will allow you to solve for the amplitude and phase.

 

[Cynapse]

"A particle moves along the x-axis. It is initially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration -0.320 m/s^2. Assume it moves with simple harmonic motion for 4.50 s and x=0 is its equilibrium position. Find its position and velocity at the end of this time interval."


x=Amplitude * cos2(pi)ft

Found f using T= radical (x/a) * 2pi = 5.769 so f= 0.1734 Hz

However, even with solving for f I'm still left with two unknowns, i.e. A and t. Please help...

This is how I've done it quickly...

x(t) = Bsin[2pi.f.t] + Acos[2pi.f.t] NB because at t=0, x=0.27 you can ignore the sin part
x(t) = Acos[2pi.f.t]
v(t) = -A.2pi.f.sin[2pi.f.t] (Diffrentiated once x(t) with respect to t)
a(t) = -A.(2pi.f)^2.cos[2pi.f.t] (Diffrentiated twice x(t) with respect to t)

x(t=0) = Acos[2pi.f.0]
x(t=0) = A (cos(0) = 0)
A = 0.27

a(t=0) = -0.27.(2pi.f)^2.cos[2pi.f.0]
a(t=0) = -0.27.(2pi.f)^2 (cos(0) = 0)
-0.320 = -0.27.(2pi.f)^2
f = sqrt[0.320/(0.27.(2.pi)^2)]
f = 0.173

x(t) = 0.27.cos[2pi.0.173.t]
x(t=4.5) = 0.27.cos[2pi.0.173.4.5]
x(t=4.5) = 0.050m

Dont quote me though its a while since I've done SHM

 

[Doc Al]

This is how I've done it quickly...

x(t) = Bsin[2pi.f.t] + Acos[2pi.f.t] NB because at t=0, x=0.27 you can ignore the sin part
x(t) = Acos[2pi.f.t]
v(t) = -A.2pi.f.sin[2pi.f.t] (Diffrentiated once x(t) with respect to t)
a(t) = -A.(2pi.f)^2.cos[2pi.f.t] (Diffrentiated twice x(t) with respect to t)

x(t=0) = Acos[2pi.f.0]
x(t=0) = A (cos(0) = 0)
A = 0.27

Sanity check: Do you think that the particle is at maximum displacement at t = 0?

Better to start with this:
$$x = A\sin(\omega t + \phi)$$