Simple Harmonic Motion and equilibrium

[Patdon10]

Homework Statement

A 100-g block hangs from a spring with k = 5.2 N/m. At t = 0 s, the block is .20 m below the equilibrium position and moving upward with a speed of 2.10 m/s. What is the block's speed when the displacement from equilibrium is .324 m?

Homework Equations

w₀=sqrt(k/m)
x(t) = Bsin(w₀*t) + Ccos(w₀*t)
v(t) = w₀Bcos(w₀*t)-w₀Csin(w₀*t)

The Attempt at a Solution

I know the answer is 1.02 m/s, and I'm not getting that at all : /

w₀=sqrt(k/m)
= sqroot (5.2/.100) = 7.2111 rad/s

x(t) = Bsin(0) + Ccos(0)
x(t) = B

v(t) = w₀Bcos(0) - w₀sin(0)
v(t) = w₀B
1.9 = 7.2111B ---> B = 0.26348

x(t) = Bsin(w₀*t) + Ccos(w₀*t)
0.297 = 0.26348*sin(7.2111*t) + (0.2)cos(7.2111*t)
0.297 = 0.03307344t + 0.19841808t
0.297 = 0.2314915203t
t = 1.28298436

v(t) = [(7.2111)(0.26348)cos(7.2111*1.28298436)] -
[(7.2111)(0.2)sin(7.2111*1.28298436)]

v(t) = 2.418364634 - 0.2318690865
v(t) = 2.31869 m/s


Can anyone see what I'm doing wrong? Thanks in advance!

 

[rl.bhat]

In the simple harmonic motion the velocity is given by

v = \omega\sqrt{(a^2 - x^2)}

You have found ω. From the given data, you can find the amplitude a. In the second case using a and x, find v.

 

[Patdon10]

Is the amplitude simply:
x = Asin(w_0*t)

I don't think that equation would work, because I don't have a time in the given. Or would the amplitude simply be what B is, because it matches the amplitude equation.

edit:
of course that wouldn't make sense either because that would make V a non-real answer.

 

[rl.bhat]

2.1 = 7.2111(a^2 - 0.2^2)^1/2

Solving this you get

a^2 = 0.1248. Next

v = 7.2111(0.1248 - 0.324^2)1/2

Solve this to find v.

 

[Patdon10]

Thanks for the response!

ok, I see now. I should have been able to figure that out from the givens. I didn't understand why you didn't have a squared on the amplitude in the equation, but I see you already added it. Why would my method not work? What am I doing wrong?

 

[rl.bhat]

simple harmonic motion is expressed as

y = Asin(ωt) or x = Acos(ωt)

The velocity at any instant is given by

v = dy/dt = Aωcos(ωt) = ω(A^2 - y^2)^1/2