Simple Harmonic Motion - Vertical Spring

[meep82817]

Homework Statement

A block with mass m =7 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.23 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.7 m/s. The block oscillates on the spring without friction.

2. Questions

After t = 0.36 s what is the speed of the block?
What is the magnitude of the maximum acceleration of the block?
At t = 0.36 s what is the magnitude of the net force on the block?

3. Equations and the attempt at a solution

I calculated k by m*g/x
w by sqrt(k/m)
A by V*sqrt(m/k) (based on energy)

For speed, I used v(t) = -w*A*sin(w*t)
For acceleration, I used amax = wˆ2*A
For force, I used F(t) = k*A*sin(w*t)

I get the feeling my mistake is on my calculation for A. Help please? I've been working on this for a while and am starting to get frustrated.

 

[Klockan3]

Can you write out what all your parameters are instead of just denoting them by w, A etc. Would make it so much easier to see what you do.

 

[meep82817]

Sure.
- k is spring constant
- A is amplitude
- w is angular frequency (rad/s)

 

[Klockan3]

A by V*sqrt(m/k) (based on energy)

What is large V here?

 

[meep82817]

Oh. My bad. I used the velocity of 4.7 m/s that the statement provides.

 

[Klockan3]

The amplitude of the oscillation gives the total energy of the spring through the use of potential energy. So E = 1/2 A²k or A = sqrt(2E/k) = sqrt(mV²/k) = V sqrt(m/k) so is the same as yours. Looking through what you got, you got your speed wrong, otherwise it looks good.