Mass hanging on a spring - find acceleration

  • Thread starter kchurchi
  • Start date
  • #1
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Homework Statement


A massless spring is hanging vertically. With no load on the spring, it has a length of 0.29 m. When a mass of 0.33 kg is hung on it, the equilibrium length is 0.92 m. At t=0, the mass (which is at the equilibrium point) is given a velocity of 4.80 m/s downward.
At t=0.70 s, what is the acceleration of the mass? (Positive for upward acceleration, negative for downward)


Homework Equations



Fspring = -k*(x(t) - xeqb)

ƩFy,mass = Fspring - m*g = -m*ay
ƩFy,mass = -k*(x(t) - xeqb) - m*g = -m*a

x(t) = A*cos(ω*t)

The Attempt at a Solution


Not sure how to go about this. I attempted to find the acceleration using the sum of the forces on the mass (which is accelerating downward) but I ran into a roadblock with finding k. I thought the equation for the position of a mass undergoing SHM might help out but I am not sure how to solve for A, because A is not equal to the equilibrium position.
 

Answers and Replies

  • #2
ehild
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First find the spring constant.


ehild
 
  • #3
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The spring constant, k, is equal to the weight force divided by the change in position.

k = (m*g)/Δx

which we are all given.

Once I've found the spring constant, now what do I do? I am confused on how to find x(t=0.7s) when I have an unknown amplitude.
 
  • #4
ehild
Homework Helper
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The mass will oscillate about the new equilibrium position. It starts SHM from equilibrium point with maximum velocity. Write up the equations for displacement, velocity, acceleration of this SHM.

ehild
 

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