Tricky spring problem. Amplitudes and AccelerationMax

  • Thread starter yigh
  • Start date
  • #1
2
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Homework Statement



A block with mass m =6.7 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.26 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.

3)After t = 0.4 s what is the speed of the block?
4)What is the magnitude of the maximum acceleration of the block?

Calculated K = 252.5384615
Oscillation Frequency (omega) = .9771168341

Homework Equations



Amplitude = vMax*sqrt(m/k)



The Attempt at a Solution


I got amplitude to be 0.7655.
I used -(omega)*Acos(omega*t) to find V for number 3. I used aMax = omega^2*t
 

Answers and Replies

  • #2
318
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I believe there is an equation that you use to find the velocity at any given point given amplitude, stretched distance, k and mass. hold on.... so its the sqrt(k/m) times the sqrt(amplitude^2 - stretched or compressed distance^2) use that
 
  • #3
318
0
oh nevemind. use 9.8m/s^2(t) to find the speed after sometime. isnt the acceleration the same?
 

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