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1.

The System In Equilibrium:

When the system is in equilibrium the total force F on the system is 0 as there is no driving force F(t).

There is no y diplacement, and therefore no velocity term and hence no drag term either.

The force of gravity and the restoring force of the spring are the only forces able to act on the static system, and by vertue of the fact that it is static, the gravity force term mg equals the restoring force F_s.

The equation describing the system is initially:

As force is the effect of an accelerated mass, and acceleration is the second derivitve of the diplacement with respect to time, this equation becomes:

The Displaced System:

When the mass is dispaced from the equilibrium position y_e (‘by pulling down’) to position y and is the released the total force on the system is not 0.

F_s depends on y in all cases; when the mass is displaced and the spring is stretched, F_s > 0.

mg opposes the total force F which is in the positive y direction.

is the drag term which also is always in the opposite direction to the total force on the system, therefore it is initially in the negative y direction in this case.

F(t) is the driving force of the sytem which is due to the release of the displaced mass, and is directed in the positive y direction due to the fact that the mass was initially ‘pulled down’ i.e. directly away from the fixed end of the spring.

The equation for this system can be based upon that derived for a suystem in equlibrium above: but now we have to consider the facts that the total force on the system now has as driving force F(t) as its main component in the positive y direction and a drag term opposing the total force which is initially in the negative y direction, making the overall equation for the system:

If we rearrange this equation, assuming a Hooke’s Law spring:

We see that it is a linear, non-homogenous ordinary differential equation, with constant co-efficients of order 2 and degree1.

Appropriate initial conditions would be:

y(0) = y_e, y(t₁) = y_1,

2.

This equation can be expressed as a system of first order differential equations as follows:

The displacement of the mass with respect to time is the velocity given by the second derivative of the displacement of the mass with respect to time is the acceleration of t he mass and can be expressed as the derivative of the velocity . Therfore the equation overall is as follows:

this second way of writing the equation is better from a computational point of view as you only have to write the code to solve a first order differential equation and apply it however many times you need to solve the problem, where as the first equation requires a pice of code to perform a single differentiation and also a second differentiation which takes more time as is shown in the programmes Q2.a and Q2.b on the disk porovided.

3.

Solving :

Use the characteristic equation y=e^rt where r is a constant:

    

Assume the differential equation has the following form where a₁, a₂ a₃, are all arbitary constants:

It then follows that:

a₂r² e^rt + a₁re^rt + a₀e^rt = 0

Which gives:

e^rt(a₂r² + a₁r + a₀)

So the solution for r is given by:

The general solution is initially:

y = C₁e^2it + C₂e^-2it

But:

e^iq
= cosq + isinq

So from the principle of superposition:

y = y₁ + y₂ = C₁(cos(2t) + isin(2t)) + C₂(cos(-2t) + isin(-2t))

And:         y = y₁ – y₂ = C₁(cos(2t) + isin(2t)) – C₂(cos(-2t) + isin(-2t))        ??????

As y(0) = 0

0 = C₁(1) + C₂(1)

i.e.

-C₁ = C₂

Unfortunately I am lost with how to rearrange all this to get the required result, which i know is that:

where C is a constant who’s value is obtained thus:        ??????????

y(0) = 0 Þ C = 0                                ??????????

4.

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9.

If moved to the moon the spring would feel a significant difference on it extension due to the change in the force of gravity. On Earth the force of gravity acting on the mass would extend the spring to an equilibrium position y_e as in the two diagrams at the start of this project.But on the moon the force of gravity is less and as a result the spring would have a different equilibrium position that would be vertically higher than that which it has on the Earth(i.e. if y_e = 0 on Earth, then on the moon y_e > 0).

On the Earth the force of gravity acting on the mass along with the drag term cause a resistance to the motion of the oscillating spring, the spring on the moon would almost be a perfect simple harmonic oscillator and go on oscillation for ever once started. However therte is gravity on the moon so the it will eventually stop oscillating but not as quickly as would be expected on the Earth, which may cause problems for apperatus that needs to be critically damped.

Also due the lack of resistance forces the motion of the mass on the sping will seem faster than if it were on the Earth as there is less to slow it down.

10.

11.

Solving:         w = 4    y(0) = 0        

Assume a homogenious solution i.e.

Assume a particular solution as f(t) = 3sin(4t) (it’s differneial family is cos, sin).

y_p = C₁sin(4t) + C₂cos(4t)

y_p‘ = 4C₁cos(4t) – 4C₂sin(4t)

y_p” = -16C₁sin(4t) – 16C₂cos(4t)

-16C₁sin(4t) – 16C₂cos(4t) +16(C₁sin(4t) + C₂cos(4t)) = 0

Therefore the general solution to the homogenious version of the equation is:

y_p = C₁sin(4t) + C₂cos(4t)

As y(0) = 0,Þ C₂ = 0, so the solution becomes:

y_p = C₁sin(4t)

Now we need to find the complimentary solution to the homogenious solution i.e.

12.

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15.

Solving:     y(0) = 0, 4, 5, 6     

y = e^rt r = constatnt

y’ = re^rt

y” = r²e^rt

As in question 3.

It then follows that:

a₂r² e^rt + a₁re^rt + a₀e^rt = 0

Which gives:

e^rt(a₂r² + a₁r + a₀)

So the solution for r is given by:

The general solution is initially:

y = C₁e^(¼+4.987i)t + C₂e^{-(¼+4.987i)t}

But:

e^iq
= cosq + isinq

So:

y₁ = C₁(cos(2t) + isin(2t)

y₂ = C₂(cos(-2t) + isin(-2t)

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