# Applications of differential equations

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Thread starter 2 months ago
#1
I am doing Q13 b.), and I am not sure what to use for my initial conditions. I know that at t=0, x=0 as one initial condition, but I'm not sure what to use for my second initial condition. When the elevator starts moving upwards with a constant speed U, the spring stretches, so the speed of the particle is 0 at t=0? Also, does the spring remain stretched through out until the elevator stops or does it eventually start moving? I think it would be the latter because there is a differential equation associated with the movement, but it's hard to imagine the motion of the spring. Help is appreciated.
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2 months ago
#2
(Original post by Physics1872)
I am doing Q13 b.), and I am not sure what to use for my initial conditions. I know that at t=0, x=0 as one initial condition, but I'm not sure what to use for my second initial condition. When the elevator starts moving upwards with a constant speed U, the spring stretches, so the speed of the particle is 0 at t=0? Also, does the spring remain stretched through out until the elevator stops or does it eventually start moving? I think it would be the latter because there is a differential equation associated with the movement, but it's hard to imagine the motion of the spring. Help is appreciated.
At , we have and at we have .

These are your two conditions.
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Thread starter 2 months ago
#3
(Original post by RDKGames)
At , we have and at we have .

These are your two conditions.
In the solution, they have used dx/dt=0 at t=0.
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2 months ago
#4
(Original post by Physics1872)
In the solution, they have used dx/dt=0 at t=0.
That's correct for a). The particle is hanging in equilibrium, then the lift starts to move for t>=0 with constant velocity. This produces an upwards force on the particle for t >= 0, but x=0 and dx//dt=0 at t=0.

In a sense it's just like releasing a stone from rest under the influence of gravity. The position and speed are initially zero but the force is applied for t>=0. The speed increases linearly and the position quadratically. The introduction of a spring in the question adds the sinusoidal (SHM) terms and the equilibrium position moves as a linear function of t because the lift speed is constant.
Last edited by mqb2766; 2 months ago
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Thread starter 2 months ago
#5
(Original post by mqb2766)
That's correct for a). The particle is hanging in equilibrium, then the lift starts to move for t>=0 with constant velocity. This produces an upwards force on the particle for t >= 0, but x=0 and dx//dt=0 at t=0.

In a sense it's just like releasing a stone from rest under the influence of gravity. The position and speed are initially zero but the force is applied for t>=0. The speed increases linearly and the position quadratically. The introduction of a spring in the question adds the sinusoidal (SHM) terms and the equilibrium position moves as a linear function of t because the lift speed is constant.
Thanks, I finally understand what is happening in this situation.
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Thread starter 2 months ago
#6
(Original post by RDKGames)
At , we have and at we have .

These are your two conditions.
Thanks for the help! I can't give rep or else I would.
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