Question

You are in charge of drinks at the graduation dinner and the people want carbonated root beer. You come up with the ingenious idea of freezing a

100 kg
block of dry ice on the end of a spring and immersing the spring-mass system in a large vat containing the root beer. You pull the dry ice down 1 meter and let it go. The dry ice melts so that the mass of dry ice, as a function of time, is given by
m(t)=100− 2
1
​
t
where
t
is in minutes. The friction coefficient is 1 and the spring constant increases as the spring gets cold, so it is given by
k(t)=2t+1
. a. Create an IVP which models this situation. b. Derive the power series solution to the IVP. c. What is the position of the dry ice after 1 minute? If your answer is approximate, explain how far off you believe you are from the true value and why. d. Sketch the graph of the motion of the dry ice for the first few minutes. Explain how you got your graph and for how many minutes you believe the graph is reasonably accurate. Explain how you came up with the number of minutes.

Answer 1

To model this situation, we can create an initial value problem (IVP) in the form of a second-order differential equation.

First, we can find the equation of motion for the spring-mass system using Newton's second law. This gives us:

F = ma = m(t) * a(t) = m(t) * d^2x/dt^2

where F is the force acting on the mass, m(t) is the mass of the dry ice at time t, a(t) is the acceleration of the mass at time t, and x(t) is the position of the mass at time t.

The force acting on the mass includes the spring force, which is given by Hooke's law:

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

where k(t) is the spring constant at time t and x(t) is the position of the mass.

The force also includes the damping force, which is given by:

Fdamping = -b * dx/dt

where b is the friction coefficient and dx/dt is the velocity of the mass.

Combining these equations, we get:

m(t) * d^2x/dt^2 = -k(t) * x(t) - b * dx/dt

This is a second-order differential equation, which we can rewrite as a system of first-order differential equations by introducing the velocity v(t) = dx/dt as a new variable:

dx/dt = v(t)

dv/dt = -(k(t)/m(t)) * x(t) - (b/m(t)) * v(t)

This is the initial value problem (IVP) that models the motion of the dry ice in the root beer.

To find the power series solution to this IVP, we can substitute the series solutions x(t) = sum(a_n * t^n) and v(t) = sum(b_n * t^n) into the system of differential equations and solve for the coefficients a_n and b_n.

To find the position of the dry ice after 1 minute, we can substitute t = 1 into the power series solution for x(t). This gives us the approximate position of the dry ice after 1 minute.

To sketch the graph of the motion of the dry ice for the first few minutes, we can use the power series solution for x(t) and v(t) to plot the position and velocity of the dry ice as a function of time. We can then use this plot to visualize the motion of the dry ice over time.

It is difficult to determine exactly how accurate the power series solution will be, as it depends on the behavior of the system and the values of the coefficients. However, the power series solution should be reasonably accurate for a small range of time around t = 0, as the series is constructed using a Taylor expansion around this point.