Solving a Differential Equation from story problem

In summary, the conversation discusses a problem involving a mass, a spring, and a differential equation. The conversation includes attempts at solving the problem and determining the position of the mass at a given time. However, there are discrepancies in the given values and equations, leading to an incorrect solution. The correct solution involves using consistent units and correcting the spring constant.
  • #1
TG3
66
0
Homework Statement

A mass weighing 2 lbs stretches a spring 6 inches. If the mass is then pulled down an additional 3 inches and then released, and there is no damping, determine the position u of the mass at ay time T according to the form Acos(wT-phi).


The attempt at a solution

The k of the spring is 2 lbs/ 6 inches, or 1/3 lbs/inches. The mass is 2lbs/32ft/sec^2, or 1/16.

From this I get the differential equation:
1/16U'' +1/3U = 0

Making U e^rt and skipping a few steps
1/16r^2+1/3 = 0
Using the quadratic:
r = 0+- 8 sqroot(1/12) (-1/12 actually, but it doesn't matter here.)

So: y = C1 cosine (8 sqroot(1/12)T) + C2 sine (8 sqroot(1/12)T)

The initial condition of y(0) ought to be 3, since the spring was stretched 3 inches, and y'(0) ought to be 0 since it was released not pushed, correct? (I'm suspicious of this step.)

If so, y(0) = C1 cosine (8 sqroot(1/12)T) + 0 (since the sine of 0 is 0).
And furthermore, y(0) = C1, so C1 =3.

y'(0) = [0] + C2 (8 sqroot(1/12)) cosine (8 sqroot(1/12)T)
y'(0) = C2 (8 sqroot (1/12))
0 = C2

So y=3cosine(8sqroot(1/12)T)

From this, translating to the form of A cosine (wT-phi) isn't hard:
A = (3^2)^.5 = 3
W = 8 sqroot(1/12)
And phi is tan^-1 (0/3) = 0

My solution is 3 cosine 8(sqroot(1/12))T.

Unfortunately, this is wrong. Where did I goof up?
 
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  • #2
TG3 said:
Homework Statement

A mass weighing 2 lbs stretches a spring 6 inches. If the mass is then pulled down an additional 3 inches and then released, and there is no damping, determine the position u of the mass at ay time T according to the form Acos(wT-phi).


The attempt at a solution

The k of the spring is 2 lbs/ 6 inches, or 1/3 lbs/inches. The mass is 2lbs/32ft/sec^2, or 1/16.

From this I get the differential equation:
1/16U'' +1/3U = 0
There are also two initial conditions. You should state them with your differential equation.
TG3 said:
Making U e^rt and skipping a few steps
1/16r^2+1/3 = 0
Using the quadratic:
r = 0+- 8 sqroot(1/12) (-1/12 actually, but it doesn't matter here.)
Well, really it does, since it makes the difference between complex solutions for r and real solutions.

In any case,
TG3 said:
So: y = C1 cosine (8 sqroot(1/12)T) + C2 sine (8 sqroot(1/12)T)
Be consistent in your variables. The dependent variable started off as u; now it's y.

TG3 said:
The initial condition of y(0) ought to be 3, since the spring was stretched 3 inches, and y'(0) ought to be 0 since it was released not pushed, correct? (I'm suspicious of this step.)
It all depends on which direction you choose to be the positive y axis. I would be more inclined to say that y(0) = -3.
TG3 said:
If so, y(0) = C1 cosine (8 sqroot(1/12)T) + 0 (since the sine of 0 is 0).
And furthermore, y(0) = C1, so C1 =3.

y'(0) = [0] + C2 (8 sqroot(1/12)) cosine (8 sqroot(1/12)T)
y'(0) = C2 (8 sqroot (1/12))
0 = C2

So y=3cosine(8sqroot(1/12)T)

From this, translating to the form of A cosine (wT-phi) isn't hard:
A = (3^2)^.5 = 3
W = 8 sqroot(1/12)
And phi is tan^-1 (0/3) = 0

My solution is 3 cosine 8(sqroot(1/12))T.

Unfortunately, this is wrong. Where did I goof up?

It could be that the book's answer has rationalized the radical. What answer does the book give?
 
  • #3
The answer given is:
.25 cos(8t)
 
  • #4
Your spring constant is incorrect, which leads to an incorrect differential equation. The mass is 1/16 slug (1 slug = 32 lb/(ft/sec)^2). Your spring constant is 2lb/6 in = 4 lb/ft. The spring constant should not be in terms of inches.

All length measures, including the initial condition, should be in terms of feet, not inches.
 

What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between an unknown function and its derivatives. It involves the use of calculus to determine the behavior of the function as the input value changes.

Why do we need to solve differential equations?

Differential equations are used to model and understand various real-world phenomena, such as population growth, chemical reactions, and motion. By solving these equations, we can make predictions and analyze the behavior of these systems.

How do you solve a differential equation from a story problem?

First, identify the variables and parameters in the problem and determine the type of differential equation (e.g. ordinary or partial). Then, use appropriate techniques such as separation of variables, substitution, or the method of undetermined coefficients to solve the equation and find the solution that satisfies the given initial conditions.

What are some common techniques used to solve differential equations?

Some common techniques include separation of variables, substitution, the method of undetermined coefficients, and the Laplace transform. Each technique is useful for different types of differential equations and may require additional steps or assumptions.

Are there any limitations to solving differential equations?

Yes, there are some limitations to solving differential equations. In some cases, it may not be possible to find a closed-form solution, and numerical methods may be needed. Additionally, solving certain types of differential equations can be very complex and may require advanced mathematical knowledge.

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