# Math/Calculus

A spring with a 4 kg mass has natural length 1 m and is maintained stretched to a length of 1.3 m by a force of 24.3 N. If the spring is compressed to a length of 0.8 m and then released with zero velocity, find the position of the mass at any time t.

Here is my work so far:

24.3/0.3 = k*0.3/0.3
k = 81
m*k" + k*x = 0
4*k" + 81*x = 0
x = +/- 4.5i
x(t) = c1*cos(4.5t) + c2*sin(4.5t)
x(0) = 0.3
x(0) = c1 -> c1 = 0.3
x'(t) = -4.5*c1*sin(4.5t)+4.5*c2*cos(4.5t)
x'(-0.2)=1.057+4.5*c2*0.6216=0
c2 = -0.378
x(t)=0.3*cos(4.5t)-0.378*sin(4.5t)

This answer is wrong. Can someone please look at my work and correct any errors. Thanks for your help.

Call the variable y the length of the spring at time t. If the oscillation is vertical and g = 9.8 m/s^2, the equilibrium position is y = 1 + Mg/k = 1.484 m
It starts out at length 0.8 m, compressed by an amount 0.684 m.
The oscillation amplitude is A = 0.684 m
y = 1.484 - 0.684 cos (4.5 t)

You have assumed the wrong amount of compression when oscillation begins
The answer will be different if g=0 or if the oscillation is in the horizontal direction.

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1. For the spring question:
The work is right but your initial conditions were incorrect.

x(0)=-0.2 --> This shows that at time t=0 the spring is compress from its natural length of 1 meter to 0.8 meters. 0.8-1=-0.2

x'(0)=0 --> This show that time t=0 the springs velocity it 0 m/s. You tried to put a distance value in x' when x' is only a measurement of velocity as a function of time.

If you use these initial conditions instead you should end up with:
x(t)= -(1/5)sin(4.5t)

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posted by Chase
2. The answer is actually -(1/5)cos(4.5t) not sin. When you plug in your initial condition for the first derivative x'(0)=0, it cancels out the C2 variable which correlates with sin. Thus leaving you with just the cos.

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posted by James
3. y``-2y`-3y=e^-2x

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